Question

A stone is thrown vertically upwards with a velocity of 10ms^-1 from the top of a tower 40m tall and it finally falls to the ground,

1. Find the time taken by the stone to reach the ground

2. After how long will it pass through the point of projection

3. Calculate the velocity when it strikes the ground. Take g = 10 ms^-2 

Answer 1

Let the stone be thrown upward with a velocity u = 10ms^-1 from the tower AB of height 40 m. Let C be the highest point reached. 

1. Let t be the total time taken by the stone to move from B to C and back to the ground.

∴ The displacement = BA = – 40 m (directed downward since displacement is measured from the initial to the final position)

a = -10 ms^-2.

From s = ut + 1/2at², we have

– 40 = 10 × t – (10t²/2)

– 40 = 10t – 5t²

– 8 = 2t - t²

t² – 2t – 8 = 0

(t – 4) (t + 2) = 0.

Or

t = + 4s or – 2s

As time cannot be negative total time taken to reach the ground = 4s.

2. For the motion from B to C and back to B, displacement = 0

u = 10ms^-1, a = g = -10ms^-2

Let t₁ be the time taken.

From s = ut + 1/2at², we have

0 = 10t₁ – (10t₁²/2) or 10₁² – 5t₁² = 0

∴ t₁ = 2s.

Hence, the stone reaches the point of projection after 2 seconds. 

3. Let ‘v’ be the velocity of the stone when it strikes the ground. For the motion from B to C and back to A, we have,

u = 10 ms^-1, a = g = -10 ms^-2, t = 4s

From, v = u + at, we have

v = 10 – 10 × 4 = – 30 ms^-1.

The -ve sign indicates that it is directed downward.