Question

A ball is thrown up with a speed of 0.5 m/s.take the value of accleration due to gravity 9.8m/s .how high will it gobefore it begain to fall?how long it will take to reach that height.

Answer 1

Here, u=0.5 m/s,  v=0, s=?,  t=?

Let us take, a=−10 m/s2 in vertical upward motion of the ball upto highest point.

Using the third equation of motion, we get
                                                                 v2=u2+2as or, s=v2−u22a          =0−0.522×(−10)=0.0125 m

Now, using the first equation of motion, we get
                                                                       v=u+at0=0.5+(−10) tor, t=0.05 s