Question

a ball is thrown upward with a speed of 49ms-1 and at same time another Ball is dropped from the top of a tower of height 100m . then after how much time they will meet each other and what is the position of the stones?

Answer 1

A ball is thrown vertically upward with speed 49 m/s , Let after t time it strikes with another body .
then,
displacement covered by ball ( S1)
S1 = ut + 1/2at²
= 49t - 1/2 × 9.8t²
= 49t - 4.9t²

second ball dropped from the top of tower of height 100m , at same time t it strikes with first ball .
then, displacemnt covered by ball (S2) ,
S2 = 1/2gt²
-S2 = - 1/2 × 9.8t²
S2 = 4.9t²

now,
striking possible only when,
S1 + S2 = 100 m
49t - 4.9t² + 4.9t² = 100
49t = 100
t = 100/49 =2.04 sec

position of first stone from below = 49×2.04 - 4.9× (2.04)² = 79.568 m

position of second ball from the top = 4.9 × (2.04)² = 20.39 m

Answer 2

Answer:

A ball is thrown vertically upward with speed 49 m/s , Let after t time it strikes with another body .

then,

displacement covered by ball ( S1)

S1 = ut + 1/2at²

= 49t - 1/2 × 9.8t²

= 49t - 4.9t²

second ball dropped from the top of tower of height 100m , at same time t it strikes with first ball .

then, displacemnt covered by ball (S2) ,

S2 = 1/2gt²

-S2 = - 1/2 × 9.8t²

S2 = 4.9t²

now,

striking possible only when,

S1 + S2 = 100 m

49t - 4.9t² + 4.9t² = 100

49t = 100

t = 100/49 =2.04 sec

position of first stone from below = 49×2.04 - 4.9× (2.04)² = 79.568 m

position of second ball from the top = 4.9 × (2.04)² = 20.39 m

Explanation: