Falling Objects| College Physics| Openstax| Problem 2.47|

(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?

SOLUTION:

Part a

Refer to Figure 2.47a.

The known values are: $t=2.35\:s;\:y=0\:m;\:v_0=+8.00\:m/s;\:a=-9.8\:m/s^2$

Based from the given values, we can use the equation $y=y_0+v_0t+\frac{1}{2}at^2$. Substituting the values, we have

$y=\left(0\:m\right)+\left(+8.00\:m/s\right)\left(2.35\:s\right)+\frac{1}{2}\left(-9.80\:m/s^2\right)\left(2.35\:s\right)^2$

$y=-8.26\:m$

Therefore, the cliff is 8.26 meters high.

Part b

Refer to Figure 2.47b.

The knowns now are: $y=0\:m;\:y_0=8.26\:m;\:v_0=-8.00\:m/s;\:a=-9.80\:m/s^2$

Based from the given values, we can use the equation $y=y_0+v_0t+\frac{1}{2}at^2$. Substituting the values, we have

$0\:m=8.26\:m+\left(-8.00\:m/s\right)t+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2$

Using the quadratic formula to solve for the value of t, we have

$t=0.717\:s$