Initial velocity using angle, distance, time

[gforrey]

Homework Statement

From the edge of a rooftop, a boy throws a stone at an angle of 25.0 degrees above the horizontal. the stone hits the ground 4.20 seconds later, 105 m away from the base of the building.(ignore air resistance). Find the initial velocity of the stone, the intial height from where the stone was launched, and the max hieght reached. I am stuck and cannot figure out which equation to use to find the inital velocity!

Homework Equations

no idea.

The Attempt at a Solution

 

[anathema]

Hello, it seems like you are trying to solve a projectile motion problem. To find the initial velocity, you can use the equation v = u + at, where v is the final velocity (which is 0 since the stone hits the ground), u is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2), and t is the time (4.20 seconds). You can rearrange the equation to solve for u, which will give you the initial velocity of the stone.

To find the initial height, you can use the equation s = ut + (1/2)at^2, where s is the displacement (in this case, it is the horizontal distance of 105 m), u is the initial velocity, t is the time, and a is the acceleration due to gravity. Rearranging the equation will give you the initial height from where the stone was launched.

To find the maximum height reached, you can use the equation v^2 = u^2 + 2as, where v is the final velocity (which is 0 at the maximum height), u is the initial velocity, a is the acceleration due to gravity, and s is the displacement (which will be the maximum height). Rearranging the equation will give you the maximum height reached by the stone.

I hope this helps! Let me know if you have any further questions.