Question

A stone is thrown from the top of a building with an initial velocity of 21.5 m/s straight upward, at an initial height of 51.7 m above the ground.

(a) Determine the time needed for the stone to reach its maximum height.

(b) Determine the maximum height.

(c) Determine the time needed for the stone to return to the height from which it was thrown, and the velocity of the stone at that instant.

(d) Determine the time needed for the stone to reach the ground.

(e) Determine the velocity and position of the stone at t = 5.85 s.

Answer 1

Solution : 1. Initial velocity u = 21.5 m/s

So the time required for maximum height

V = u + at

Here v = 0 and a = g = -9.81 m/s2

So t = u / g = 21.5 / 9.81 = 2.19 seconds.

2. Maximum height

V² =u² + 2gH at max height V = 0

So H = 21.5 x 21.5 / 2× 9.81

H = 23.56 meters that is 23.6 meters from initial point. From ground 51.7+23.6 =75.3 meter

3.total time to reach at initial height =

2 x time of going up = 2. X 2.19 = 4.38 seconds

And velocity would be same as thrown with hence

V = 21.5 m/sec

4. For total.time of journey

We need to find the time taken by stone to cover 51.7 metre distance .

Final velocity of the stone on the ground

V² = u² + 2gh

So V² = 21.5 x 21.5 + 2 x 9.81 x 51.7 m

So v = 38.4 meter/ seconds

Now time.

V = u + at

So t= v - u / a

So t = 38.4 - 21.5 / 9.81 = 1.72 seconds

Ao total time = 4.38 + 1.72 = 6.10 seconds .

5. For position we use

S = ut + 0.5 at²

Lets take u = 21.5 m/s and a = -9.81 m/s2

So s = 21.5 x5.85 - 0.5 × 9.81 * 5.85*5.85

So s = -42.1 meter it means the stone is at 51.7 - 42.1

= 9.6 meters above from the ground .

Now speed v = u + at = 35.9 m/ s