Posted by
**Josh** on
.

A man stands on the roof of a 19.0 m-tall building and throws a rock with a velocity of magnitude 24.0 m/s at an angle of 38.0 degrees above the horizontal. You can ignore air resistance.

Calculate the maximum height above the roof reached by the rock.

got it.... 11.1 m

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

got it.... 30.8 m/s

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

DON'T KNOW

an stands on the roof of a 19.0 m-tall building and throws a rock with a velocity of magnitude 24.0 m/s at an angle of 38.0 degrees above the horizontal. You can ignore air resistance.

Calculate the maximum height above the roof reached by the rock.

got it.... 11.1 m

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

got it.... 30.8 m/s

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

DON'T KNOW

You know the initial horizontal velocity (24*cos38). You can figure the time in air (a little tricky, using the vertical distance equation

hinitial=hfinal + 24Sin38*t - 1/2 9.8 t^2, solve this quadratic for time, use the quadratic formula)

Now, knowing horizontal velocity and time in air, you can easily calculate horizontal distance.

calculating the horizontal distance is this:

x=V0t * t

so ur initial velocity is 24cos(38) * t

t is the time it took to reach the ground

hi, how did u get the velocity of the rock when it hits the ground?