Question

You throw a rock straight up from the edge of a cliff. It leaves your hand at time t = 0 moving at 13.0 m/s. Air resistance can be neglected. Find both times at which the rock is 4.00 m above where it left your hand. Enter your answers in ascending order separated by a comma. Express your answer in seconds.

Answer 1

Answer:

0.36s, 2.3s

Explanation:

Let gravitational acceleration g = 9.81 m/s2. And let the throwing point as the ground 0 for the upward motion. The equation of motion for the rock leaving your hand can be written as the following:

$s = v_0t + gt^2/2$

where s = 4 m is the position at 4m above your hand. $v_0 = 13 m/s$ is the initial speed of the rock when it leaves your hand. g = -9.81m/s2 is the deceleration because it's in the downward direction. And t it the time(s) it take to get to 4m, which we are looking for

$4 = 13t - 9.81t^2/2$

$4.905 t^2 - 13t + 4 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{13\pm \sqrt{(-13)^2 - 4*(4.905)*(4)}}{2*(4.905)}$

$t= \frac{13\pm9.51}{9.81}$

t = 2.3 or t = 0.36

Answer 2

Explanation:

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