## Question

A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed of 200 meters per second.

(a) How much time elapses before the bullet hits the ground?

(b) How far does the bullet travel horizontally?

### Answer

**The answer to a is 0.553 seconds**

**While for b, the answer is 110.6 meters**

#### Explanation

**Data**

We will need to identify our data from the question

The height, h = 1.5 m

Horizontal velocity, v = 200 m/s

Initial velocity, u = 0

Acceleration due to gravity, g = 9.8 ms^{-2}

##### Answer to a

To find out how much time elapses before the bullet hits the ground, we will follow the method below:

**Unknown **

Elapsed time before the bullet hits the ground, t = ?

**Formula**

We will apply the formula,

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

Since the initial velocity, u = 0

We can rewrite h = ut + (1/2)at^{2} into

h = (1/2)at^{2}

We can now make t subject of the formula

at^{2} = 2h

t = √(2h/a)

but we are dealing with the force of gravity, which implies a = g

Thus

t = √(2h/g)

**Therefore, to find the elapsed time (t), we will use the formula t = √(2h/g)**

###### Solution

To find the elapsed time, we will now substitute our formula with our data

t = √(2h/g) = √((2 x 1.5) / 9.8)

Which implies that

t = √(3 / 9.8) = √0.306 = 0.553 seconds

**Therefore, the elapsed time before the bullet hits the ground is 0.553 seconds**

##### Answer to b

**Unknown**

Distance traveled by the bullet horizontally, s = ?

**Formula**

Distance traveled by the bullet horizontally, s = horizontal velocity (v) x elapsed time (t)

###### Solution

We will insert our data into the above formula

s = v x t = 200 x 0.553 = 110.6 m

**Therefore, the distance traveled by the bullet horizontally is 110.6 meters**

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