A object is thrown upward from a building that is 70m high. I am trying to solve

Question

A object is thrown upward from a building that is 70m high.
I am trying to solve for the distance upward to its maximum height and i am making a silly mistake. Initial velocity is 12m/s, final velocity is zero. with -g. the time is 1.22 seconds up. When I try to solve for the distance I am doing something wrong. I multiply 12m/s x 1.22 seconds and get 15m. This is the wrong equation, why? Instead the answer comes from the equation (0m^2/s^2)=12m^2/s^2 + 2(-9.8m/s^2) x (change in distance)
and -12^2/2 x (-9.8) = 7.4
7.4m is the correct distance for an object thrown up from a 70m building with an initial velocity of 12m/s. Why does only one of these equations work???

Explanation / Answer

Given that the initial velocity is u = 12 m/s
Final velocity at maximum height h is v = 0
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Apply kinematic equation v^2 - u^2 = 2as for
this motion, we get
v^2 - u^2 = 2(-g)h
0 - (12 m/s)^2 = 2(-9.8 m/s^2)h
h = (12 m/s)^2 / 2(9.8 m/s^2)
h = 7.34 m
This is the maximum height reached by the body
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In the case of pojected body the body having acceleration
that acceleration is equal to acceleration due to gravity.
This acceleration due to gravity constant throughout the motion
So we must apply kinematic equations for the vertical motion.

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