Question

A rock is tossed straight up from ground level with a speed of 22 m/s. When it returns, it falls into a hole 9.0 m deep. a. What is the rock’s velocity as it hits the bottom of the hole? b. How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Answer 1

Answer:

(a) $v_{f}=26.08m/s$

(b) $t=4.906seconds$

Explanation:

We will take ground level as origin and upward is positive direction

the givens are

$y_{i}=0\\y_{f}=-9.0m\\ v_{i}=22m/s$

Part (a)

To find the final velocity we use the kinematic equation

So

$(v_{f})^{2} =(v_{i})^{2} +2a(y_{f}-y_{i})\\(v_{f})^{2}=(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)\\v_{f}=\sqrt{(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)} \\v_{f}=26.08m/s$

Part (b)

To find time of rock trip until it touches the ground we will use simple kinematic equation or simple motion equation

$v_{f}=v_{i}+at\\t=\frac{v_{f}-v_{i}}{a}\\ t=\frac{(-26.08m/s)-(22m/s)}{-9.8m/s^{2} }\\t=4.906seconds$

Notice that we substituted vf with negative sign because its direction is downwards