Question

Shyanne drops a rock from a hill at initial height of 975 feet above ground level. How many seconds after the rock is released will it hit the ground below

61 seconds
7.8secdons
5.2 seconds
11.6seconds

Answer 1

61 second I would say but what is the weight of the rock?

Answer 2

Option B. 7.8 seconds

Explanation:

Shyanne drops a rock from a hill at initial height of 975 feet above the ground level.

We have to calculate the time taken by rock to hit the ground.

As we know the formula of motion under gravity is

h = ut +
`(1)/(2)gt^(2)`

Here h = initial height = 975 feet

u = initial velocity = 0

t = time taken by the rock to hit the ground

g = 32.174
`\frac{\text{Feet}}{\text{Second}^(2)}`

Now we plug in these values in the formula

975 = 0 +
`(1)/(2)(32.174)(t)^(2)`

975 = 16.087t²

t² =
`(975)/(16.087)`

t =
`√(60.60)`

t = 7.78 ≈ 7.8 seconds

Option B. 7.8 seconds is the answer.