Question

Therefore, time for which the parachutist remain inair,
20
t = +13= 15-86 s
7
Problem 1.23. A ball is allowed to fall from the top of a
tower 200 m high. At the same instant, another ball is thrown
vertically upwards from the bottom of the tower with a
velocity of 40 m s-1. When and where the two balls meet?
Sol. Suppose that the two balls meet after a timet at a height
h above the ground.
For the ball dropped from the top: The ball will cover a
distance equal to (200 - h) m.
Therefore, u = 0; S = (200 - h) m;a= g = 9.8 m 5-2
1 2​

Answer 1

Answer:

after 5 second

77.5 from the ground

Explanation:

1/2*9.8*25

49*5/2

245/2

122.5

200-122.5

77.5

Answer 2

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hope it helps