Question

A ball thrown vertically upwards returns to the hand in 4.8s. (a.) How high did the ball go? (b.) What was the ball's initial speed? (c.) At what time(s) is the ball's speed equal to 19.6 m/s?

Answer 1

Given:-

• Final velocity (v) = 0m/s

• Time taken (t) = 4.8s

• Acceleration due to gravity (g) = 9.8m/s²

To Find:-

• (a) height attained by the ball (h)

• (b) Initial velocity (u)

• (c) Time taken by ball to attain speed 19.6m/s.

Solution:-

(b)

Firstly we calculate the final velocity of the ball

Using 1st equation of motion

→ v = u+at

Substitute the value we get

→ 0 = u + (-9.8) ×4.8

→ -u = -47.04

→ u = 47.04 m/s

∴ The initial velocity of the ball is 47.04m/s

Now (a)

calculating the height attained by the ball

Using 3rd equation of motion

→ v² = u² + 2as

→ 0² = (47.04)² + 2×(-9.8) × s

→ 0 = 2212.76 + (19.6) ×s

→ -2212.76 = -19.6 ×s

→ s = -2212.76/-19.6

→ s = 112.89 ≈ 113 m

∴ The height attained by the ball is 113 Metre.

(c)

Calculating the time taken to attained the speed of 19.6m/s

Using 1st equation of motion we get

→ v = u+at

→ 0 = 19.6 + (-9.6) × t

→ -19.6 = -9.8×t

→ t = -19.6/9.8

→ t = 2s

∴ The time taken by the ball to attain the speed of 19.6m/s is 2 second.