Question

A rock is thrown upward with the velocity of 27 meters per second from the top of a 43 meter high cliff, and it misses the cliff on the way back down. When will the rock be 8 meters from ground level?

Answer 1

The rock will be 8 meters from ground level in 6.6 seconds.

Solution:

$h(t) = \frac{-1}{2}gt^2 + v_0t + h_0$

The value of gravity, $g = 9.8 m/s^2$

$h(t) = \frac{-1}{2}\times{9.8}t^2 + 27t + 43$

To find: The time when h(t) = 8 m from ground level

$\Rightarrow-4.9t^2 + 27t + 43 = 8$

$\Rightarrow-4.9t^2 + 27t + 35 = 0$

Using quadratic formula: $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\text { when } a x^{2}+b x+c=0$

Here, the unknown value x is considered as t. On substituting the values of a as -4.9; b as 27 and c as 35 we get,

$t=\frac{-(27) \pm \sqrt{(27)^{2}-4(-4.9)(35)}}{2(-4.9)}$

$t=\frac{-27 \pm \sqrt{729+686}}{-9.8}$

$t=\frac{-27 \pm \sqrt{1415}}{-9.8}$

Square root of 1415 is 37.61648 which is approximately 37.6

$t=\frac{-27 \pm 37.6}{-9.8}$

$t=\frac{-27+37.6}{-9.8} \text{ or } t=\frac{-27-37.6}{-9.8}$

$t=\frac{10.6}{-9.8} \text{ or } t=\frac{-64.6}{-9.8}$

$t=-1.08163 \text{ or } t=6.5918367$

On ignoring the negative value since time will be of positive value we get,

$t=6.5918367\approx t=6.6 \text{ seconds }$