A man stands on the roof of a building of height 17.0 m and throws a rock with a velocity of magnitude 33.0 m/s at an angle of 33.0 ∘ above the horizontal. You can ignore air resistance.(A) Calculate the maximum height above the roof reached by the rock. (B) Calculate the magnitude of the velocity of the rock just before it strikes the ground.(C) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

(A). The maximum height above the roof reached by the rock is 16.48 m.

(B). The velocity of the rock before hitting the ground is 37.6 m/s.

(C). The horizontal distance from the base of the building to the point where the rock strikes the ground is 122.7 m.

Explanation:

Given that,

Height = 17.0 m

Velocity = 33.0 m/s

Angle = 33.0°

(A). We need to calculate the maximum height above the roof reached by the rock

Using formula of height

Put the value into the formula

(B). We need to calculate the time

Using equation of motion

Put the value into the formula

We need to calculate the  magnitude of the velocity of the rock just before it strikes the ground

Using equation of motion

The velocity of the rock before hitting the ground is

(C). We need to calculate the horizontal distance from the base of the building to the point where the rock strikes the ground

Using equation of motion

Put the value into the formula

Hence, (A). The maximum height above the roof reached by the rock is 16.48 m.

(B). The velocity of the rock before hitting the ground is 37.6 m/s.

(C). The horizontal distance from the base of the building to the point where the rock strikes the ground is 122.7 m.

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.