Question

A person throws balls into air one after the other at an interval of 1 s the next ball is thrown when the velocity of the first ball thrown is 0 .to what height will the ball rise

Answer 1

Dear Student

Here $$\begin{gathered} timet=1s \\ finalvelocityv=0m/s \\ so \\ v=u+gt \\ 0=u+(-9.8\times 1) \\ u=9.8m/s \\ soforheight \\ h=ut+\frac{1}{2}g{t}^2 \\ h=9.8\times 1+\frac{1}{2}(-9.8)\times 1\times 1 \\ h=9.8-4.9 \\ h=4.9m \end{gathered}$$
Regards
Sandeep Kumar