Question

When a ball is thrown up vertically with velocity $v_0$, it reaches a maximum height of 'h'. If one wishes to triple the maximum height then the ball should be thrown with velocity?

• A. v_0
• B. 3v_0
• C. 9v_0
• D. 3/2 v_0

Answer 1

The provided options do not contain the correct answer, which is $\sqrt{3}v_0$.

Answer 2

The maximum height $h$ reached by an object thrown vertically upwards with an initial velocity $v_0$ is given by $h = \frac{v_0^2}{2g}$. If we want to triple the maximum height to $h' = 3h$, and the new initial velocity is $v'$, then $h' = \frac{(v')^2}{2g}$. Substituting $h' = 3h$: $3h = \frac{(v')^2}{2g}$ Substitute $h = \frac{v_0^2}{2g}$: $3 \left(\frac{v_0^2}{2g}\right) = \frac{(v')^2}{2g}$ $3v_0^2 = (v')^2$ $v' = \sqrt{3}v_0$. Since $\sqrt{3}v_0$ is not among the options, none of the provided choices are correct.