Question

A man stands on the roof of a building of height 16.8m and throws a rock with a velocity of magnitude 30.6m/s at an angle of 29.3? above the horizontal. You can ignore air resistance.

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

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

Answer 1

there are two different answers to the horizontal distance, depending on the direction the man threw the rock - up or down at the 29.3 degree angle, impact velocity (magnitude and angle) will be the same whether the rock was thrown upwards or downwards, assuming no air friction.

first the rock thrown down calculation (assuming no air friction)

Vxo = Horizontal velocity = 30.6m/s * cos 29.2 = 26.68m/s
Vyo = initial Vertical velocity = 30.6m/s * sin 29.3 = 14.975m/s down

to calculate the time it takes for the rock to hit the ground.
x = Vyo * t + 1/2 * a * t^2
16.8m = 14.975m/s * t + 1/2 * 9.81 * t^2
t=0.872 s

the final vertical speed is Vyo + a * t = 14.975 + 9.81 * 0.872 = 23.52 m/s
velocity at impact = sqrt(Vx^2 + Vy^2) = sqrt(26.68^2 + 23.52^2) = 35.56m/s
angle at impact = arctan(Vy/Vx) = arctan(23.52/26.68) = 41.3 degrees (down from horizontal)

the horizontal distance the rock hits from the building = Vxo * t = (26.68 * 0.872) = 23.26m