A steel ball is dropped from a building\'s roof and passes a window, taking 0.13

Question

A steel ball is dropped from a building's roof and passes a window, taking 0.13 s to fall from the top to the bottom of the window, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.13 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2.17 s. How tall is the building?

I know I need to do something with h = 1/2 (g)t^2, but I keep getting so mixed up with what to do with all the given times.

Explanation / Answer

We know the equation of motion,

s=ut+0.5*gt^2

for the flight from the top of the building to top of the window,

s1 = 0*t + 0.5*g*t1^2 <-----here initial speed=u=0, time taken fom top of building to top of window = t1

Also, from top of building to bottom of window,

s2 = 0*t+0.5*g*t2^2

So, distance between these two, s' = s2-s1 = 1.2 m <----given in the question

So, 0.5*g*(t2^2-t1^2) = 1.2

So, 0.5*g*(t2+t1)*(t2-t1) = 1.2 <----using the identity = a^2-b^2 = (a+b)*(a-b)

Now, if you carefully observe, t2-t1 = 0.13 s <----given from the data

t1+t2 = time if remains up above the window(lets call it Ta) + 0.13

So,

0.5*g*(Ta+0.13)*0.13 = 1.2 <----taking g = 9.8m/s2

so, Ta = 1.75 s

So, the total flight can be summed up by :

total time = total time above the window+time in between+time below the window

or, T = Ta+Tbe+Tbl

Also, Tbe = 2*0.13

So, T = 1.75+2*0.13+2.17 = 4.18 s

We now see that this is the total flight time from top of the building to bottom and then from bottom to top,

So, for top of building to bottom only time taken = 4.18/2 = 2.09s

so, using the same equation,

s= 0*t +0.5*9.8*2.09^2 = 21.4 m <-----------------answer

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