Question

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it strike the ground?

Answer 1

`t = 1.82`

Step-by-step explanation:

Given

`u = 7.70m/s` -- initial velocity

`s = 30.2m` --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

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

Where

`g = 9,8m/s^2`

So, we have:

`30.2 = 7.70t + (1)/(2) * 9.8 * t^2`

`30.2 = 7.70t + 4.9 * t^2`

Subtract 30.2 from both sides

`30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2`

`0 = 7.70t + 4.9 * t^2 - 30.2`

`0 = 7.70t + 4.9t^2 - 30.2`

`7.70t + 4.9t^2 - 30.2 = 0`

`4.9t^2 + 7.70t - 30.2 = 0`

Solve using quadratic formula:

`t = (-b\±√(b^2 - 4ac))/(2a)`

Where

`a = 4.9;\ b = 7.70;\ c = -30.2`

`t = (-7.70\±√(7.70^2 - 4*4.9*-30.2))/(2*4.9)`

`t = (-7.70\±√(651.21))/(9.8)`

`t = (-7.70\±25.52)/(9.8)`

Split the expression

`t = (-7.70+25.52)/(9.8)` or
`t = (-7.70-25.52)/(9.8)`

`t = (17.82)/(9.8)` or
`t = -(33.22)/(9.8)`

Time can't be negative; So, we have:

`t = (17.82)/(9.8)`

`t = 1.82`

Hence, the time to hit the ground is 1.82 seconds