Question

A rock is dropped from the top of a vertical cliff and takes 3 s to reach the ground below the cliff. A second rock is thrown vertically from the top of the cliff, and it takes this rock 2 s to reach the ground below the cliff from the time it is released. With what velocity was the second rock thrown, assuming no air resistance?

Answer 1

The rock is thrown with a velocity equal to 12.25
`(m)/(s)`

Step-by-step explanation:

Steps

• Find the vertical distance from the first scenario given.
• Use the vertical distance(s) from the first scenario, to solve for the velocity in the second scenario.

Scenario One:

s =
`(1)/(2) gt^(2)`

where s = distance, g = 9.81
`(m)/(s^(2) )`, t = time

This gives us s = 44.1m

Scenario Two:

From Newton's equation of motion, we have
`s = ut + (1)/(2) at^(2)`

where a = acceleration due to gravity ( 9.8m/s^2)

s = displacement of the rock equal to 44.1m

t = time taken to complete this downward motion, equal to 2 seconds

u = initial velocity at which the rock is thrown vertically

Therefore, substituting the above into Newton's equation gives

44.1m = 2u + 0.5(-9.8 x
`2^(2)`)

After simplifying, we get

44.1m = 2u +
`(-39.2)/(2)`

2u = 44.1 - 19.6, Therefore, u = 12.25
`(m)/(s)`

Note:

In a vertical velocity situation, when an object is dropped, the initial velocity is zero(0) when in a free fall because it has left contact with whatever held it. But when it is thrown, it has the exact same initial speed in free fall as it did before it was released.