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?

To calculate the velocity of the rock just before it strikes the ground, you can use the fact that the horizontal component of velocity remains constant throughout the motion (since there is no horizontal acceleration).

First, find the time it takes for the rock to reach the ground. You can use the vertical motion equation:

h_initial = h_final + v_initial * sin(theta) * t - 1/2 * g * t^2

where:
h_initial is the initial height (19.0 m)
h_final is the final height (0 m, since the rock hits the ground)
v_initial is the initial velocity (24.0 m/s)
theta is the launch angle (38.0 degrees)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time it takes for the rock to reach the ground (unknown)

Rearrange the equation to solve for t:

0 = 19.0 + 24.0 * sin(38) * t - 1/2 * 9.8 * t^2

This is a quadratic equation, so you can solve it using the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = -4.9, b = 24 * sin(38), and c = 19. Solve for t using the quadratic formula.

Once you have the time, you can use the horizontal velocity (v_initial * cos(theta)) and the time to calculate the horizontal distance using the equation:

horizontal distance = horizontal velocity * time taken to reach the ground

Substitute the known values to calculate the horizontal distance.