Question

When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 metres.

1. What was the initial speed of the ball?
2. How much time is taken by the ball to reach the highest point? (g = 10 ms⁻²)

Answer 1

(a) v (final velocity) = 0 m/s

u (initial velocity) = ?

a (acceleration due to gravity i.e., g) = 10 m/s²

S (distance) = 5 m

v² − u² = 2aS

(0)² − (u)² = 2 × 10 × 5

−(u)² = 2 × 10 × 5

−(u)² = 100

u =  −10 m/s

u =  −10 m/s

The initial velocity was negative, i.e., −10 m/s, as it was velocity against gravity because the ball was thrown upwards. This can be neglected.

(b) s (distance) = 5 m

u (initial velocity) = 10 m/s

a (acceleration due to gravity, i.e., g) = 10 m/s²

t (time) = ?

`S = ut +1/2 g t^2`

⇒ `5 = 10 xx t + 1/2 xx 10 xx t^2`

⇒ 5 = 10 t + 5 t²

⇒ t² + 2t − 1 = 0

⇒ t = `(sqrt(2) - 1)` s or `(-sqrt(2) - 1)` s

Since , time cannot be negative, the required answer is `(sqrt(2) - 1)` s