Question

A rock is thrown off a cliff with a speed of 5 m/s downward. How far will it fall after 7 seconds have elapsed?y = yo + vo t + ½ a t 2 g = 9.8 m/s2 (downward)

Answer 1

Given data:

* The initial velocity of the rock is 5 m/s.

* The time taken by the rock is 7 seconds.

* The acceleration of the rock is,

`g=9.8ms^(-2)`

Solution:

By the kinematics equation,

The distance traveled by the rock in the vertical direction is,

`y=y_o+v_ot+(1)/(2)gt^2`

where y_o is the initial distance of the rock, y is the final distance after the time t, v_o is the initial velocity, and t is the time taken to reach the final distance,

Substituting the known values,

`\begin{gathered} y=0+5*7+(1)/(2)*9.8*7^2 \\ y=35+240.1 \\ y=275.1\text{ m} \end{gathered}`

Thus, the distance traveled by the rock in 7 seconds is 275.1 meter.