A golf ball is thrown straight up from the edge of the roof of a building. A sec

Question

A golf ball is thrown straight up from the edge of the roof of a building. A second golf ball is dropped from the roof a time of 1.10 s later. You may ignore air resistance.

1.If the height of the building is 20.3 m , what must the initial speed be of the first ball if both are to hit the ground at the same time? The initial speed is 8.89m/s I got this one.

2.But I don't know how to do the second one: Consider the same situation, but now let the initial speed v0 of the first ball be given and treat the height h of the building as an unknown. What must the height of the building be for both balls to reach the ground at the same time for v0 = 9.00 m/s .

Explanation / Answer

1- Let the speed of first ball be v
The height, H which it reaches above the building = (v^2/2g) = H
The time, T to go to the top for ball 1 = v/g = T
Let the height of the building be h. we have
v/g + sq rt [2*(H+h)/g] = 1.10 + sq rt[2h/g] = 1.10 + sq rt[(2*20.3)/9.8} = 3.135, Multiplying by g gives,
sq rt [2*g*(H+h)] = 3.135*9.8 - v , Squaring we get
2gH + 2gh = (30.72 -v)^2 or
v^2 +2gh = (30.72 -v)^2 or
2gh = (30.72 - v)^2 - v^2 = 30.72*(30.72 - 2*v) = 2*9.8*20.3 = 397.88 or
2*30.72*v = 545.8384 or v = 8.89 m/s

2. Same equations will hold and derivations will work; v will be repalced by Vo = 9.0
Similarly we have,
Vo/g + sq rt [2*(H+h)/g] = 1.10 + sq rt[2h/g]; multiplying by g we get
Vo + sq rt[2*g*(H+h)] = 1.10*g + sq rt[2gh]
we are suppoed to find out h, but 2gh = V^2,say then if we know V, the velocity with which the second ball reaches ground we can know h. replacing variable h by vaiable V and putting all data, we have
9.0 + sq rt[Vo^2+V^2] = 1.10*9.8 + V = 10.78 +V or
sq rt[Vo^2+V^2] = 1.78 + V or
Vo^2 +V^2 = 1.78^2 +V^2 + 2*1.78*V or
9.0^2 - 1.78^2 = 3.56*V = 77.8316 or V = 21.86
So h = (21.86^2)/(2*9.8)
= 24.4 m

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