Answer to Question #4361 in Mechanics | Relativity for Nate

Question

A playground is on the flat roof of a city school, 6.1 m above the street below (see figure). The vertical wall of the building is h = 7.50 m high, forming a 1.4-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of θ = 53.0°; above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.

(a) Find the speed at which the ball was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.)

Answer 1

Equation of motion of the ballx(t) =v₀ cos(teta t) y(t)=v₀ sin(teta t) -gt^2/2
We know that at the moment t=2.2 the ball is 7.5 highover the ground.

7.5=2.2v₀ sin(teta)-4.84g/2= 2.2v₀ sin(53)-4.84g/2=1.757v₀-23.716 => v₀=17.77 m/s