Question

A cricket ball thrown vertically upwards with an acceleration of 10m/s2 reaches a maximum height of 5m Find the initial speed of the ball and the time taken to reach the Highest point?

Answer 1

v=0m/s

h=5m

g=9.8m/s^2

(a) v^2=u^2+2gh

0^2=u^2+2(-9.8×5)

0=u^2-2×49

0=u^2-98

98=u^2

√98=u

9.8m/s=u


(b) v=u+gt

0=9.8+9.8t

-9.8=9.8t

-9.8÷9.8=t

1s=t


hope it helps
mark as brainliest✌️✌️
follow me ❤️
love love

Answer 2

time taken by the ball is 1 sec and the initial speed is 10 m/s

s=ut+ 1/2 at^2

5= 0+1/2×10× t^2. ( u =0 because it is at maximum height )

5=5t^2

t^2=1

t=1

now,

v=u+at

0= u+10

u=-10 m/s ( - sign indicate that ball is moving upward)

hope it may help you

please mark me as brainliest!!!!!!!