Answer to Question #4361 in Mechanics | Relativity for Nate

Question #4361
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.)
1
Expert's answer
2011-10-06T13:00:10-0400
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

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