Physics
hannahgonzales8
4

A ball thrown horizontall at 22.2 m/s from the roof of a building lands 36.0m from the base of the building. How tall is the building?

+0
(1) Answers
bethanychaplin

Velocity of the ball (v) = 22.2 m/s Distance from the base of the building where the ball falls (d) = 36.0m We know that, Velocity (v) = distance (d) / time (t) From the above equation, Time (t) = distance (d) / velocity (v) Substituting the values for distance and velocity in the above equation, Time = 36.0/22.2 = 1.62 seconds When the ball is falling down, it accelerates due to gravity This acceleration (a) = 9.8 m/s² Now we need to find the height of the building. The equation of motion for an object with constant acceleration is, s = ut + 1/2 at² Where, s = vertical displacement, which is height of the building in our case u = initial velocity of the ball = 0 Substituting all the above values in the equation of motion, s = (0 × 1.62) + (1/2 × 9.8 × 1.62²) s = 0 + (25.72/2) = 12.86 m = 12.9 m which is the height of the building Therefore the building is 12.9m tall

Add answer