Question

Two stones are projected from the top of a tower 100m high, each with a velocity of 20 m/s. One is projected vertically upward and the other vertically downwards. Calculate the time each stone takes to reach the ground and the velocity with which it strikes the ground?

Answer 1

The stone which is thrown vertical upwards.

initial velocity, u = 20m/s

height of stone from the ground, h = 100m

time taken by stone to become rest , t = u/g = 20/10 = 2sec

height reached by ball from the top of tower, x = u²/2g = (20)²/2(10) = 20m,

now stone starts to fall at height S = (h + x) = 100 + 20 = 120m, downward with initial velocity , u = 0.

so, 120 = 0. t' + 1/2 (10) t'²

or, t'² = 24 => t' = 2√6 sec

hence, total time taken to reach the ground by stone , T = t + t' = (2 + 2√6)sec

now, velocity of stone with it strikes the ground, v = √(u² + 2gh)

= √{(20)² + 2(10)(100)}

= √(2400) m/s

The stone which is thrown vertically downwards.

initial velocity, u = 20m/s

height of stone from the ground, h = 100m

time taken to reach the ground by the stone, t

so using formula, s = ut + 1/2 at²

or, 100 = 20t + 1/2(10)t²

or, 10 = 2t + 1/2t²

or, t² + 4t - 20 = 0

or, t = {-4 ± √(16 + 80)}/2

= (-2 ± 2√6)sec but t can't be negative so, t = (-2 + 2√6)sec

velocity of stone with its strikes the ground, v = √(u² + 2gh) = √(2400) m/s

Answer 2

Explanation:

The stone which is thrown vertical upwards.

initial velocity, u = 20m/s

height of stone from the ground, h = 100m

time taken by stone to become rest,t= u/g=20/10 = 2sec height reached by ball from the top of

tower, x = u²/2g = (20)²/2(10) = 20m,

now stone starts to fall at height S = (h +x) = 100+ 20 = 120m, downward with initial velocity, u=0.

so, 120 = 0. t' + 1/2 (10) t¹²

or, t¹² = 24 => t' = 2:√6 sec hence, total time taken to reach the

ground by stone, T=t+t' = (2+2√6)sec

now, velocity of stone with it strikes the ground, v = √(u² + 2gh)

= {(20) + 2(10)(100)} = √(2400) m/s =

The stone which is thrown vertically

downwards.

initial velocity, u = 20m/s

height of stone from the ground, h = 100m

time taken to reach the ground by the stone, t

so using formula, s = ut + 1/2 at²

or, 100 = 20t + 1/2(10)t²

or, 10 = 2t+1/2t2

or, t² + 4t-20=0

or, t = {-4 ± √(16+80)}/2

= (-2 ± 2-√6)sec but t can't be negative so, t = (-2 + 2√6)sec

velocity of stone with its strikes the ground, v = √(u² + 2gh) = √(2400) m/s