Trajectory of kicked ball

Question

A ball is kicked from the ground with a vertical velocity of $10 m/s$; how long does it take the ball to reach maximum height above ground?

I used equation $v=u+at$ but rearranged it and used the $9.8$ as there was nothing else given. Is the answer 1.026?

Answer 1

Yes it is correct indeed

• $v=v_0-gt=0 \implies t=\frac{v_0}{g}$

and the maximum height is

• $h=v_0 t-\frac12gt^2=\frac{v_0^2}{g}-\frac12g\frac{v_0^2}{g^2}=\frac12\frac{v_0^2}{g}$