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Find the time when the ball is 5.00 m below the roof railing. The motion involves constant
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Jan 10, 2025
Find the time when the ball is 5.00 m below the roof railing. The motion involves constant acceleration due to gravity.
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find the time when the ball is 5.0 M
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below the roof
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rail the motion of involves constant
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acceleration due to gravity so we are to
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find the time when the ball is 5.0 M
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below the roof rail and the motion
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involves constant acceleration due to
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gravity which is G therefore our data is
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initial position y equal 0 initial
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velocity B of yal to+ 15.0 m/s
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and final position y = - 5.0 M
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acceleration due to gravity which is gal
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9.8 m/s squ so this is the data that we
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are supposed to use to arrive at our
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answer now the key equation the position
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of the ball as a function of time is
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given by y = y plus b of Y t+ half a y
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of t² this is a an equation of motion
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and secondly since a of y isalus g
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therefore we have y we will now
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substitute the equation as y = y + V of
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Y tus GT 2 because a of Yus G therefore
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replace it
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by that substituting y notal 0 we will
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now arrive at yal to B of YT minus GT s
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because y will disappear because it is
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now is zero and when we rearrange into
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standard quad uh quadratic form we'll
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now rearrange it we now have half GT squ
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because we are now rearranging that
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previous equation and we'll arrive at
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half
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gtus b not of YT + y = z why because we
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want to arrive we want to make sure that
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we collect like terms and everything
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equal to Z and at the same time this
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equation can also be written as G / 2
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tus B YT + y =
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0 when we rearrange the equation because
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we want to make it in form of quadratic
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equation therefore now coefficients of
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the quadratic equations are we now need
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the coefficients of quadratic equation
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which is AAL G / 2 which is 9.8 / 2
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because G is = 9.8 we now have 4.9 balus
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B of Y which is- 15.0 and C is = to Y
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which is - 5.0 therefore the quadratic
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equation now becomes
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4.9 0 t² - 15.0 t - 5.0 = 0 because the
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previous quadratic equation we now
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rearrange it by substituting a b c into
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our previous
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equation so now the next thing we need
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to do after rearrange this we now look
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at AB
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C the use quadratic formula tus B plus
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orus root of B S - 4 a c 2 a and we will
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now solve for t i remember that we
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already have a b and c so we now
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substitute it into this formula and
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after substituting
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this ABC into the quadratic formula we
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will now arrive
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at T equals to whatever we have here
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which is minus so so so so so and now we
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will still simplify it to arrive at the
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second equation that you are seeing here
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which is 15.0 plus or minus this we will
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now continue to simplify it and remember
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that- 15 s =
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225.0 and down the reason why we have 4
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9.8 is because we have 2 * 4.9 and it
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will give
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for
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9.8 so we'll continue to simplify this
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equation tal
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to 15 plus plus or minus therefore by
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simplif by means of simplification we
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will now consider it as plus and then
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consider it as minus therefore in this
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case we will now say T1 and T2 so to
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make T1 or to after
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simplification once we arrive at 15.0
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plus orus 1797 all over 9.8 that is when
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you will begin to see it from the
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perspective of positive or negative
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which is T1 for positive and T2 for
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negative that is after you have done all
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the simplification so we will now
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consider it in form of T1 and T2 then T1
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we will now use FL and T2 will use
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negative so for T1 we arrive at 15 plus
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17.9 7 all over 9.8 and it will give us
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3.3 seconds 6 seconds which is positive
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for T1 and for2 which is negative we now
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have 15.9 instead of because it's plus
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or minus now minus 17.9 7 all over 9.8
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and it will give us 0 - 0.3 seconds
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therefore 41 = 3.36 seconds this is the
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time when the ball is 5.0 M below the
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rail after being released while for T2
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which isus 0.3 seconds 30 seconds this
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is not fiscally valid because it implies
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that time before the ball was released
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or it is talking about the time when the
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ball was not released while for T1 will
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have time when the ball was released
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which is 5.0 M below the railing after
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be released and this is what you have
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