Question

1. (a) calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) how long would it take to reach the ground if it is thrown straight down with the same speed?

Explanation:

Step1: Use kinematic equation for part (a)

The kinematic equation $y - y_0=v_0t-\frac{1}{2}gt^2$. Here $y - y_0=-h$ (taking down - ward as negative, $h$ is the height of the cliff), $v_0 = 8.00\ m/s$, $t = 2.35\ s$ and $g=9.8\ m/s^2$.
So, $-h=v_0t-\frac{1}{2}gt^2$.
Substitute the values: $-h=(8.00\ m/s)\times2.35\ s-\frac{1}{2}\times9.8\ m/s^2\times(2.35\ s)^2$.
First, calculate $(8.00\ m/s)\times2.35\ s = 18.8\ m$ and $\frac{1}{2}\times9.8\ m/s^2\times(2.35\ s)^2=\frac{1}{2}\times9.8\times5.5225\ m = 27.06025\ m$.
Then $-h=18.8\ m - 27.06025\ m=- 8.26025\ m$. So, $h = 8.26\ m$.

Step2: Use kinematic equation for part (b)

The kinematic equation is still $y - y_0=v_0t-\frac{1}{2}gt^2$, but now $v_0=- 8.00\ m/s$ (down - ward direction), $y - y_0=-h=-8.26025\ m$ and $g = 9.8\ m/s^2$.
So, $-8.26025=-8.00t-\frac{1}{2}\times9.8t^2$.
Rearrange to get a quadratic equation $4.9t^2 + 8.00t-8.26025 = 0$.
The quadratic formula for $ax^2+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$. Here $a = 4.9$, $b = 8.00$ and $c=-8.26025$.
First, calculate the discriminant $\Delta=b^2 - 4ac=(8.00)^2-4\times4.9\times(-8.26025)=64+161.7009 = 225.7009$.
Then $t=\frac{-8.00\pm\sqrt{225.7009}}{9.8}=\frac{-8.00\pm15.0233}{9.8}$.
We take the positive root $t=\frac{-8.00 + 15.0233}{9.8}=\frac{7.0233}{9.8}=0.72\ s$.

Answer:

(a) $8.26\ m$
(b) $0.72\ s$