Question

A student drops a rock from rest at a distance h above the ground such that the rock hits the ground at time t0. At what distance above the ground should the rock be dropped such that it hits the ground at a time 2t0 after it is released from rest?

Answer 1

Answer: 4h

Step-by-step explanation:

This is a classic "free fall" problem. The following equations describe the motion of a free falling mass:

$y(t) = y_{0} + v_{0}.t - \frac{1}{2}.g.t^{2}\\v(t) = v_{0}-g.t$

where $y_{0}$ is the initial position of the mass, $v_{0}$ its initial speed and $g$ is the acceleration due to gravity.

In this case, $y_{0} = h$ and $v_{0} = 0$

Replacing in the above mentioned equations we obtain:

$y(t) = h - \frac{1}{2}.g.t^{2}$

$v(t) = -g.t$

At $t = t_{0}$ we know that $y(t=t_{0}) = 0$

So

$y(t=t_{0}) = h -\frac{1}{2}.g.t_{0}^2 = 0\\\\h = \frac{1}{2}.g.t_{0}^2$

Now lets find the distance $h'$ the rock should be dropped from such that it hits the ground ($y=0$) at $t = 2.t_{0}$

$y(t=2.t_{0}) = h' - \frac{1}{2}.g.(2.t_{0})^2 = 0\\h' = 2.g.t_{0}^2 = 4.h$

Here we have the answer.

Answer 2

Building and solving a quadratic equation, it is found that the object should be dropped from a height of $4h_0$.

Considering the gravity and that the object was dropped from rest, the height of the object after t seconds, dropped from a height of $h_0$, is given by:

$h(t) = -4.9t^2 + h_0$

It hits the ground when $h(t) = 0$, thus:

$4.9t^2 = h_0$

$t = \sqrt{\frac{h_0}{4.9}}$

We want the time to be multiplied by 2, thus, considering the square root, the height should be multiplied by 4, that is, the object should be dropped from a height of $4h_0$.

A similar problem is given at https://brainly.com/question/17127383