Question

3. A boulder rolls with speed of 3.5 m/s off a cliff. It hits the ground 2.25 m from the base ofthe ledge. A) How high is the ledge? B) How long did it take the boulder to fall to the bottomof the cliff?DrawingVerticalHorizontal

Answer 1

Given data

*The given distance from the base of the ledge is R = 2.25 m

*The given speed is v = 3.5 m/s

The diagram is given below

(a)

Let (h) be the height of the edge

The formula for the distance from the base of the ledge is given as

$\begin{gathered} R=v\times t \\ R=v\times\sqrt[]{\frac{2h}{g}} \\ h=\frac{R^2\times g}{2v^2} \end{gathered}$

Substitute the known values in the above expression as

$\begin{gathered} h=\frac{(2.25)^2\times9.8}{2\times(3.5)^2} \\ =2.025\text{ m} \end{gathered}$

Hence, the height of the ledge is h = 2.025 m

(b)

The formula for the time taken by the boulder to fall to the bottom of the cliff is given as

$t=\sqrt[]{\frac{2h}{g}}$

Substitute the known values in the above expression as

$\begin{gathered} t=\sqrt[]{\frac{2\times2.025}{9.8}} \\ =0.642\text{ s} \end{gathered}$

Hence, the time taken by the boulder to fall to the bottom of the cliff is t = 0.642 s