Question

A ball is thrown up with a speed of 15 m/s. how high will it go before it begins to fall?
(g=,9.8m/s square)​

Answer 1

Answer:

11.48 m (approx.)

Explanation:

Given:

u (initial velocity) = 15 m/s

v (final velocity) = 0 m/s  (Since the object will stop before falling down)

a (acceleration) = -9.8 m/s²

(We are taking acceleration as a negative value because the object is decelerating)

To find:

s (displacement)

Formula:

v² = u² + 2as

0 = 225 + (-19.6s)

19.6s = 225

s = 225/19.6 = 11.4795918367 ≈ 11.48 m

Hence, the object will reach a height of approximately 11.48 m before it begins to fall.