Question

A ball thrown vertically upwards with a velocity of 10m/s.Calculate the height attain by ball and time taken by it to reach the hghest point if the acceleration of ball is -10m/s2 during its motion?(5m, 1s)

Answer 1

Since,v=0,u=10,g=-10,to find t and h

Answer : v^2=u^2-2gS
0=-100-2(-10)h
0=-100+20h
100/20=h
Therefore,h =5m
v=u-gt
0=-10+10t
10/10=t
Therefore,t=1s

Answer 2

The height attained by ball and time taken by it to reach the highest point is 5 meters and 1 second respectively.

Explanation:

Initial speed of the ball, u = 10 m/s

At the maximum height, the final speed of the ball, v = 0

Acceleration of the ball, $a=-10\ m/s^2$

Let t is the time taken by the ball. Using first equation of motion to find it as :

$v=u+at\\\\-u=at\\\\t=\dfrac{-u}{a}\\\\t=\dfrac{-10}{-10}\\\\t=1\ s$

Let h is the height attained by the ball. Using third equation of motion to find it as :

$v^2-u^2=2ah\\\\-u^2=2ah\\\\h=\dfrac{-u^2}{2a}\\\\h=\dfrac{-(10)^2}{2\times (-10)}\\\\h=5\ m$

So, the height attained by ball and time taken by it to reach the highest point is 5 meters and 1 second respectively.

Learn more,

Kinematics

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