## Solutions to the Problems of Projectile Motion

Here are 11 solved problems of projectile motion to help you understand how to tackle any related question:

### Problem 1

A stone is shot out from a catapult with an initial velocity of 65 meters per second at an elevation of 60 degrees. Find

b. Maximum Height attained

c. The range

**Data:**

Initial velocity, u = 65 m/s

The angle of elevation, θ = 60^{0}

Gravitational acceleration, g = 10 m/s^{2}

a. To find the time of the flight, we will apply the following methods:

**Unknown**

Time of flight, T = ?

**Formula**

The formula for solving a problem on time of flight, T is (2usinθ) / g

#### Solution

We will now insert our data into the above formula

Since T = (2usinθ) / g

T = (2 x 65 x sin60^{0}) / 10 = (130 x 0.866) / 10 = 112.58 / 10 = 11.258 = 11.3 seconds

**Therefore, the time of flight is 11.3 seconds**

b. Here is how to calculate the maximum height

**Unknown**

Maximum Height, H_{max} = ?

**Formula**

The formula for calculating the maximum height is

Maximum height, H_{max} = (u^{2}sin^{2}θ) / 2g

#### Solution

We can now substitute our formula with the data

The maximum height, H_{max} = (65^{2}sin^{2}60^{0}) / 2 x 10

Therefore,

H_{max} = (4225 x (sin60^{0})^{2}) / 2 x 10

And the above equation will give us

H_{max} = (4225 x 0.75) / 20

Hence

H_{max} = 3168.75 / 20 = 158.4375 m

Our maximum height can be approximated into

H_{max} = 158 m

**Therefore, our maximum height is 158 meters**.

c. We will find the range

**Unknown **

The range, R = ?

**Formula**

Here is the formula that will help us find the range

The range, R = (u^{2}sin2θ) / g

#### Solution

The range, R = (65^{2}sin2(60^{0}) / g

And the above expression will give us

R = (4225sin120^{0}) / 10 = (4225 x 0.866) / 10 = 3658.85 / 10 = 365.885 m

**Therefore, the range of the stone shot out of the catapult is 365.885 meters**.

### Problem 2

A bullet is fired horizontally with a velocity of 40 meters per second from the top of a building 50 meters high. How far from the feet of the building will the bullet be assumed to touch the ground? (Take the gravitational acceleration, g = 10 m/s^{2}).

**Data**

Horizontal velocity = Initial velocity = u_{x} = 40 m/s

Height of the building, H = 50 m

Time of flight, T = √ (2H / g)

Hence,

T = √ (2x 50 / 10) = √ (100 / 10) = √ 10 = 3.162277 = 3 seconds

The range, R = uT = 40 x 3 = 120 m

**Therefore, the bullet will cover 120 meters from the foot of the building to the ground. **

### Problem 3

A tennis ball is thrown vertically upwards from the front with a velocity of 50 m/s. Calculate

a. The maximum height reached

b. The time it takes to reach the maximum height

c. The time of the flight

**Data**

Initial velocity, u = 50 m/s

a. Here is how to find the maximum height of the tennis ball

**Formula**

The formula for finding the maximum height is

H_{max} = u^{2} / 2g

**Solution**

After inserting our data into the formula H_{max} = u^{2} / 2g, we will have

Maximum height, H_{max} = (50)^{2} / 2 x 10 = 2,500 / 20 = 125 m

**Therefore, the maximum height reached by the tennis ball is 125 meters**

b. The time taken to reach the maximum height can be calculated this way:

**Unknown**

The time it takes to reach the maximum height, t = ?

**Formula**

The formula to apply is t = u / g

**Solution**

We will apply our data to the formula t = u / g

Time, t = 50 / 10 = 5 seconds

c. Time of flight will be

T = 2u / g = (2 x 50) / 10 = 100 / 10 = 10 seconds.

### Problem 4

A bullet is fired from a point making an angle of 36 degrees to the horizontal. The initial velocity of the bullet is 43 m/s. Find

a. The greatest height reached

b. The time taken to reach the maximum height

[Take gravitational acceleration, g = 9.8 m/s^{2}]

**Data**

The angle made, θ = 36^{0}

Initial velocity, u = 43 m/s

Gravitational acceleration, g = 9.8 m/s^{2}

a. We can calculate the maximum height, H_{max} this way

**Unknown**

The maximum height, H_{max} = ?

**Formula**

Maximum height, H_{max} = (u^{2}sin^{2}θ) / 2g

**Solution**

Apply the data into H_{max} = (u^{2}sin^{2}θ) / 2g to obtain

H_{max} = (43^{2}sin^{2}36^{0}) / 2 x 9.8

Thus

The maximum height, H_{max} = (1849 x 0.34549) / 19.6 = 638.8 / 19.6 = 32.59 = 33 m

Therefore, the maximum height reached is 33 meters.

b. The time of flight will be

Time of flight, T = usinθ / g = 43sin36^{0} / 9.8 = (43 x 0.59) / 9.8 = 25.37 / 9.8 = 2.589 s

**Therefore, the time of flight is 2.6 seconds**

### Problem 5

A projectile is fired with an initial velocity of 100 m/s at an angle of 30 degrees with the horizontal. Calculate:

a. The time of flight

b. The maximum height attained

c. The range

**Data**

The initial velocity of the projectile, u = 100 m/s

Angle of inclination, θ = 30^{0}

Gravitational acceleration, g = 10 m/s^{2}

#### Solution a

To find the time of flight, we will follow the method below:

**Formula**

we will apply

Time of flight, T = (2usinθ) / g

We will now put our data into the formula

T = (2 x 100 x sin30^{0}) / g = (200 x 0.5) / 10 = 100 /10 = 10 s

**Therefore, the time of flight is10 seconds**

#### Solution b

To find the maximum height attained, we will use

**Formula**

Maximum height, H_{max} = (u^{2}sin^{2}θ) / 2g

We can now substitute our formula with our data

H_{max} = (100^{2 }x sin^{2}30^{0}) / (2 x 10)

H_{max} = (10,000 x (0.5)^{2}) / 20 = (10,000 x 0.25) / 20 = 2,500 / 20 = 125 m

**Therefore, the maximum height attained by the projectile is 125 meters **

#### Solution c

Now, we are going to calculate the range

**Formula**

The range, R = (u^{2}sin2θ) / g

After inserting our data into the above formula, we will get

R = (100^{2}sin2 x 30^{0}) / 10 = (10,000 x sin60^{0}) / 10 = (10,000 x 0.866) / 10

And we will have

R = 8660 / 10 = 866 m

**Therefore, our range is eight hundred and sixty-six (866) meters.**

### Problem 6

A bomber on a military mission is flying horizontally at a height of 3000 meters above the ground at 60 kilometers per minute. It drops a bomb on a target on the ground. Determine the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released.

**Data:**

The horizontal velocity of the bomber aircraft = 60 km/min = (60 x 1000) / 60 = 60,000 / 60 = 1000 m/s

Initial velocity, u = 0

Gravitational acceleration, g = 10 m/s^{2}

Height above the ground, h = 3000 m

**Formula**

The first formula is

h = ut + (1/2)gt^{2} to find t

The second formula

Distance covered in time t, s = horizontal velocity x time to find s

The third formula is

tanθ = s / h to find the angle of inclination (θ)

#### Solution

**We will need to insert our data into the first formula to find t**

h = ut + (1/2)gt^{2}

Which can be rewritten as

3000 = (1/2) x 10 x t^{2} [Remember that u = 0]

By making t subject of the formula, we will have

(3000 x 2) / 10 = t^{2}

t^{2} = 600

We can now take the square root of both sides to obtain

t = √600

And our final answer for the time will be

Time, t = 24.5 seconds

**We will apply our second formula**

Distance, s = horizontal velocity x time = 1000 x 24.5 = 24,500 meters

**We will now proceed with our third formula **

tanθ = s / h = 24,500 / 3000 = 8.2

Thus,

tanθ = 8.2

Hence, we divide both sides by tan to obtain a tan inverse of 8.2

θ = tan^{-1}8.2 = 83.047^{0}

We can now approximate our answer into

θ = 83^{0}

**Therefore, the angle of inclination is eighty-three (83) degrees.**

### Problem 7

A ball is projected horizontally from the top of a hill with a velocity of 30 meters per second. If it reaches the ground 5 seconds later. The height of the ball is

**Data:**

The initial velocity of the ball, u = 0

Time it takes to reach the ground, t = 5 s

Gravitational acceleration, g = 10 m/s^{2}

**Formula**

Maximum height of the ball, h = ut + (1/2)gt^{2}

#### Solution

To solve the problem, insert your data into the above formula

h = ut + (1/2)gt^{2} = 0 x 5 + (1/2) x 10 x 5^{2} = (1/2) x 10 x 25 = (1/2) x 250 = 250 / 2 = 125 meter

**Therefore, the height of the hill is 125 meters**

### Problem 8

A stone is projected at an angle of 60 degrees and an initial velocity of 20 meters per second. Determine the time of flight.

**Data:**

The angle of inclination of the stone, θ = 60^{0}

Initial velocity of the stone, u = 20 m/s

Gravitational acceleration, g = 10 ms^{-2}

**Unknown**

Time of flight, T = ?

**Formula**

The formula for calculating the time of flight, T = (2usinθ) / g

#### Solution

We will put our data into the formula T = (2usinθ) / g

Now,

T = (2usinθ) / g = T = (2 x 20 x sin60^{0}) / 10 = (40 x 0.866) / 10 = 34.64 / 10 = 3.464 s

**Therefore, the time of flight is 3.46 seconds.**

### Problem 9

A body is projected upward at an angle of 30 degrees with the horizontal at an initial speed of 200 meters per second. In how many seconds will it reach the ground? How far from the point of projection will it strike?

**Data**

Angle of projection, θ = 200 m/s

Horizontal initial speed, u = 200 m/s

**Unknown**

The question wants us to solve

Number of seconds for the body to reach ground = Time of flight, T = ?

We were also asked to find out how far from the point of projection will the body strike the ground. Hence, we are going to find the range.

**Formula**

First formula to find the time of flight, T = (2usinθ) / g

Second formula to find the range, R = (u^{2}sin2θ) / g

#### Solution

To find the time of flight, we will insert our data into the first formula T = (2usinθ) / g

Time of flight, T = (2usinθ) / g = (2 x 200 x sin30^{0}) / 10

Hence

T = (400 x 0.5) / 10 = 200 / 10 = 20 s

**Therefore, the time of flight is 20 seconds.**

By applying the second formula R = (u^{2}sin2θ) / g and inserting our data, we will have

R = (u^{2}sin2θ) / g = (200^{2} x sin (2×30^{0})) / 10

Which implies that

R = (40,000 x sin60^{0}) / 10 = (40,000 x 0.866) / 10

Thus,

The range, R = 34,640 / 10

Hence

R = 3464 m

**Therefore, the range covered is 3,464 meters **

### Problem 10

A tennis ball projected at an angle θ attains a range of R = 78 meters. If the velocity imparted to the ball by the racket is 30 meters per second, calculate the angle θ.

**Data **

Range, R = 78 m

The velocity imparted to the ball by racket = initial velocity, u = 30 m/s

Gravitational acceleration, g = 10 m/s^{2}

**Unknown**

The angle θ = ?

**Formula**

We will use the formula R = (u^{2}sin2θ) / g

#### Solution

Substitute your formula with the data

78 = (30^{2}sin2θ) / 10

Hence

780 = (900sin2θ)

Sin2θ = 780 / 900

Thus,

Sin2θ = 0.867

After dividing both sides by sin, will now have

2θ = sin^{-1}0.867

Which is equal to

2θ = 60.11^{0}

After approximating 60.11^{0} to 60^{0}. Therefore,

θ = 60^{0} / 2 = 30^{0}

**Therefore, the angle θ is 30 degrees**

### Problem 11

An aeroplane, flying in a straight line at a constant height of 500 meters with a speed of 200 meters per second drops a food package. The package takes a time t to reach the ground and travels a horizontal distance d in doing so. Taking g as 10 meters per second square, and ignoring air resistance, what are the values of t and d?

**Data**

The height, h = 500 m

Speed of the aeroplane = 200 m/s

Gravitational acceleration, g = 10 ms^{-2}

**Unknown**

The time it takes the package to reach the ground, t = ?

Distance covered by the package to reach the ground, d = ?

**Formula**

**First formula:** We will apply t = √2h/g to find time (t)

**Second Formula:** We will use s = speed of the aeroplane x time it takes the package to drop on the ground [where s = distance covered by the package to reach the ground]

#### Solution

By applying our first formula t = √2h/g and inserting our data, we will obtain

t = √((2×500) / 10) = √1000/10 = √100 = 10 seconds

**Therefore, the time it takes for the package to reach the ground is 10 seconds. **

We will now apply the second formula to calculate the distance covered

Remember that s = speed of the aeroplane x time

Thus,

s = 200 x 10 = 2000 m

**Therefore, the distance covered by the package to reach the ground is 2000 meters**

*You may also like to read:*

How to Calculate Maximum height

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