Playground Balls

A playground is on the flat roof of a city school, 6.00 m above the street below. The vertical wall of the bui

A playplace is on the flat roof of a city school, 6.00 m above the street below. The vertical wall of the building is 7.00 m high, forming a 1.00 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 26.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.

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*2.2^2
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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Part V in series Elementary Playground Balls.m4v