a rock is thrown problem....?

Question

I don't know if this is a solvable question or not but here's the question.

A rock is thrown directly upward from the edge of the roof of a building that is 56.3 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Take the acceleration due to gravity to have magnitude 9.80 m/s^2 and neglect air resistance. With what speed was the rock thrown?

it just seems like i'm not given enough information. any suggestions on how to figure this out.

Answer 1

Sure you are.

Under constant acceleration A with initial velocity V:

distance = VT + (1/2) AT^2

In this case, you have two distances, the one going up and the one going down. You also have two times, the one going up and the one going down.

The two distances are connected by the height of the building; the two times are connected by their sum.

You can get rid of the VT term by recognizing that it takes the same amount of time to go from the roof to the maximum height as it takes to go from the maximum height back down to the roof. So you can assume a V of 0.

This gives you two equations in two unknowns. Solve for the two times (up and down) and the distance above the roof.

Then, go back and use the time going up and the acceleration due to gravity to get the initial velocity.