Question

13. A ball is thrown upwards with a speed of 39.2 m/s. Calculate (a) the maximum height it reaches, and (b) the time taken in reaching the maximum height.

Answer 1

Answers please tel

Answer 2

U=39.2 a=-10m/s^2 as accelertion is opposite to gravity at maximum height v=0
Apply V^2-U^2=2as
-(39.2)^2=2*-10*s
-20s=-1536.64
s=76.832 meter
v-u=at
-39.2=-10t
t=3.92 seconds