Question

A rock is thrown directly upward from the edge of the roof of a building that is 66.2 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Neglect any effects of air resistance. With what speed was the rock thrown?

Answer 1

For the vertical position of the rock, we use the equation for free fall,

$y(t) = y_{0} + ut - \frac{1}{2} gt^2$

Here, $y(t)$ is the height of projectile at any time, t, $y_{0}$ is the initial height, u is initial speed, t is time of flight.

Given, $y_{0} = 66.2 m$, $t = 4 s$ and $g =9.8 \ m/s^2$.

We put, y(t)=0, because we have to calculate the speed at the rock thrown.

Substituting the given values, we get

$0 = 66.2 \ m + u \times 4 \ s - \frac{1}{2} (9.8 m/s^2) ( 4 \ s)^2 \\\\ u = \frac{78.4 - 66.2}{4} =3.05 \ m/s$.

Thus, the speed was the rock thrown 3.05  m/s.

Answer 2

The initial speed of the rock when it was thrown upward is 3.05 m/s.

Initial speed of the rock

The initial speed of the rock is calculated as follows;

s = ut - ¹/₂gt²

where;

• u is the speed of the rock when it was thrown

-66.2 = 4u - ¹/₂(9.8)(4²)

-66.2 = 4u - 78.4

78.4 - 66.2 = 4u

12.2 = 4u

u = 12.2/4

u = 3.05 m/s

Thus, the initial speed of the rock when it was thrown upward is 3.05 m/s.

Learn more about initial speed here: https://brainly.com/question/24493758