A playground is on the flat roof of a city school, 5.4 m above the street below

Question

A playground is on the flat roof of a city school, 5.4 m above the street below (see figure). The vertical wall of the building is h = 6.90 m high, to form a 1.5-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 ? = 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.


m/s

(b) Find the vertical distance by which the ball clears the wall.


m

(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

Explanation / Answer

Let us assume , 6.00 m above the street below and building is 7.00 m high, forming a 1.00 m high railing set the equation of motion in the vertical as y=0 at the surface of the playground for the ball y(t)=-6+v0*sin(53)*t-.5*9.81*t^2 we are given that 26=v0*cos(53)*2.20 solve for v0 19.637 m/s check with y(t) to see if the ball is above the wall y(2.2)=-6+19.637*sin(53)*2.2-.5*9.81*2… y(2.2)=4.762 m, which is 3.762 m above the wall. Let's find t for when y(t)=0 There will be two roots, use the greater root. The smaller value of t is when the ball reaches the plane of the playground on the ascent. t=2.753 seconds In that time the ball has traveled x(2.753)=35.5 m or 35.5-26 from the wall 9.5 m

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