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 1.19 s later. You may ignore air resistance. a)If the height of the building is 20.6 m , what must the initial speed be of the first ball if both are to hit the ground at the same time? b)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 = 8.65 m/s .

Explanation / Answer

a) for 2 nd ball
h = Vot + 0.5 gt^2
20.6 = 0 + 0.5 * 9.81 * t^2
t = 2.05s
time taken by 1st ball = 2.05 + 1.19 = 3.24 s
- 20.6 = Vo* 3.24- 0.5 * 9.81 * 3.24^2
Vo = 9.53 m/s

b) 2nd ball
h = ut + 0.5 gt^2
h = 0 + 0.5 * 9.81 * t1^2
t1^2 = 2h/g
for 1st ball,
time taken by 1st ball t2 = t1 + 1.19 = rt [2h/g] + 1.19
- h = 8.65 * t2 - 0.5 * g * t2^2
- h = 8.65 { rt [2h/g] + 1.1} - 0.5 g * { rt [2h/g] + 1.9}^2
- h = 8.65 rt [2h/g] + 9.515 - 0.5g [ 2h/g + 2.38rt [2h/g] + 1.42 ]
- h = 8.65 rt [2h/g] + 9.515 - h - 1.19g rt [2h/g] - 6.96
3 rt [2h/g] = 2.55
rt [2h/g] = 0.85
2h/g = 0.72
h = 3.55 m

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