A ball is thrown up at a speed of 4.0 m/s. Find the maximum height reached by the ball. Take `g=10 m//s^2`.

Explanation

Let us take vertically upward direction as the positive Y-axis. We have `u=4.0 m/s and a=-10m/s^2`. At the highest point the velocity becomes zero.Using the formula.
`v^2=u^2+2ay`,
`0=(4.0 m/s)^2+2(-10 m/s^2)y` ltbrlt or, `y=(16 m^2/s^2)/(30 m/s^2)=0.80 m`.