A ball is thrown upwards to a roof of 122.5 m away, after the first ball is going upward for 2 seconds, a second ball is thrown straight after it. Find the initial velocity of second ball so that both balls touches the roof at same time?( please read the question carefully)
Answers
Answer:
Let ball A is thrown upward at t = 0. It takes time T to reach the ground. Ball B is dropped from the roof 1.00 s later. If both the balls reach the ground simultaneously, the flying time for ball B will be (T- 1) s.
The displacement of both the balls is (= -20 m).
Using y = y
0
+ ut + (1/2)at
2
initially for both balls are at y
0
= 20 and when both reaches ground where y = 0.
For ball A: 0 = 20 + v
0
T + (1/2)(-10)T
2
...(i)
For ball B: 0 = 20 + (1/2)(-10)(T−1)
2
...(ii)
From Eqs. (i) and (ii),
v
0
T = 5(2T - 1) ...(iii)
From Eqs. (i) and (iii), we get
T
2
- 2T - 3 = 0 ⇒ (T + 1)(T - 3) = 0
T = 3s
From Eqs.(iii) v
0
= 25/3 m/s
Let ball A reaches maximum height at time t' after throwing. For ball A at maximum height, the velocity will be zero.
Using v = u + at
0 =(25/3) - 10 ×t
′
⇒t
′
=5/6 s
For maximum height reached by ball A,
Using v
2
=u
2
+2as
0 = (
3
25
)
2
- 2 x 10 x h'
h' = 125/36 m
Hence, maximum height reach by ball A
H = 20 + (125/36) = 845/36 m
Explanation:
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