Question

A ball is thrown straight up from the ground at 15 m/s. At the same time, another ball is dropped from rest from a height of 30 m. If g = 10 m/s², at what height above the ground do they meet?

पर्याय

• 5 m

• 10 m

• 15 m

• 20 m

Answer 1

10 m

Explanation:

Using s = ut + \[\frac {1}{2}\]gt², the stone moves upward with u = 15 m/s and downward acceleration 10, and the ball falls from 30 m with u = 0. Solving gives meeting time 2 s and the stone’s height s = 15 × 2 − \[\frac {1}{2}\] × 10 × 2² = 10 m above the ground.