# Chapter 2 - Motion Along a Straight Line - Problems: 45a

The ball should be thrown upwards with the speed of $31 ms^{-1}$.

#### Work Step by Step

Step-1: We need to throw a ball with velocity $v_0 ms^{-1}$ to cover a distance of $50m$ against acceleration due to gravity, which equals $-9.8ms^{-2}$. We know that when a ball is at its maximum height, its velocity at that instant is $0ms^{-1}$, that is, $v=0ms^{-1}$. Step-2: For an ascending ball, $$v^2=v_0^2-2ax$$ $$(0)^2=v_0^2-2\times (9.8) \times 50$$ $$\implies v_0=\sqrt{980}=31.30ms^{-1} \approx 31 ms^{-1}$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.