If ball A is thrown up with half the speed of B, it will not go as high, therefore will fall to the ground quicker. Using the suvat equation $s=v_{0}t+\frac{1}{2}at^2$

For A, calculate when $s=0$ to find when the ball reached the ground $$0=v_{0}t-4.5t^2$$ Solve for t to get $t=0$ and $t=\frac{v_{0}}{4.5}$

Do the same for B: $$0=2v_{0}t-4.5t^2$$ $t=0$ and $t=\frac{2v_{0}}{4.5}$

This shows that it takes twice as long for B to reach the ground

If ball A is thrown up with half the speed of B, it will not go as high, therefore will fall to the ground quicker. Using the suvat equation $s=v_{0}t+\frac{1}{2}at^2$

For A, calculate when $s=0$ to find when the ball reached the ground $$0=v_{0}t-4.5t_A^2$$ Solve for $t_A$ to get $t_A=0$ and $t_A=\frac{v_{0}}{4.5}$

Do the same for B: $$0=2v_{0}t_B-4.5t_B^2$$ $t_B=0$ and $t_B=\frac{2v_{0}}{4.5}$

This shows that it takes twice as long for B to reach the ground.

To find the heights reached by each ball use the equation $v^2=v_0^2+2as$. For A this will give: $$0=v_0^2-19.6s_A$$ $$s_A=\frac{v_0^2}{19.6}$$ The same can be done for B to give $s_B=\frac{4v_0^2}{19.6}$.This shows that B will reach a height 4 times greater than A.

If ball A is thrown up with half the speed of B, it will not go as high, therefore will fall to the ground quicker.

If ball A is thrown up with half the speed of B, it will not go as high, therefore will fall to the ground quicker. Using the suvat equation $s=v_{0}t+\frac{1}{2}at^2$

For A, calculate when $s=0$ to find when the ball reached the ground $$0=v_{0}t-4.5t^2$$ Solve for t to get $t=0$ and $t=\frac{v_{0}}{4.5}$

Do the same for B: $$0=2v_{0}t-4.5t^2$$ $t=0$ and $t=\frac{2v_{0}}{4.5}$

This shows that it takes twice as long for B to reach the ground