Question 40: Motion in 1D

Question

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $$\frac{h}{3}$$ while going up and coming down respectively.

Accepted Answer

A ball is thrown vertically upward with a certain velocity, reaching a maximum height $ h $. We need to find the ratio of the times it is at height $ \frac{h}{3} $ while ascending and descending, respectively.
The initial velocity of the ball $ v $ can be given by:
$ v = \sqrt{2gh} $
When the ball is at height $ \frac{h}{3} $, the equation of motion is:
$ \frac{h}{3} = \sqrt{2gh} \cdot t - \frac{1}{2}gt^2 $
Rearranging the equation, we get a quadratic equation in terms of $ t $:
$ \frac{g}{2} t^2 - \sqrt{2gh} \cdot t + \frac{h}{3} = 0 $
The ratio of the times taken to ascend and descend to the height $ \frac{h}{3} $ can be found using the quadratic formula solution for $ t $, considering the corresponding velocities:
$ \frac{t_1}{t_2} = \frac{\sqrt{2gh} + \sqrt{2gh - \frac{2gh}{3}} }{\sqrt{2gh} - \sqrt{2gh - \frac{2gh}{3}}} $
Simplifying the terms inside the fraction:
$ = \frac{\sqrt{2} + \frac{2}{\sqrt{3}}}{\sqrt{2} - \frac{2}{\sqrt{3}} } = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} $

Answer

$$\frac{\sqrt{2}-1}{\sqrt{2}+1}$$

Answer

$$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$

Answer

$$\frac{\sqrt{3}-1}{\sqrt{3}+1}$$

Answer

$$\frac{1}{3}$$

Motion in 1D

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.

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