2.)A playground is on the flat roof of a city school, 5.7 m above the street bel
ID: 583720 • Letter: 2
Question
2.)A playground is on the flat roof of a city school, 5.7 m above the street below (see figure). The vertical wall of the building is h = 6.80 m high, to form a 1.1-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. ___ m
Explanation / Answer
Horizontal component of ball's velocity is
24 m/2.20 s = 10.91 m/s
a) So the ball's initial speed was
10.91 m/s divided by cos(53 degrees) = 18.127 m/s
b) y = 18.127 m/s t sin(53 degrees) - (1/2) g t^2
At t = 2.20 s,
y = 14.477 m times 2.2 - (4.9 m/s^2) (2.2 s)^2
= 8.133 m
so the ball clears the 6.80-meter wall by 1.33 m
c) ball lands when y = 5.7 m on the way down.
5.7 = 14.477 t - (4.9) t^2
0 = 4.9t^2 - 14.477 t + 5.7
t = (14.477 plus or minus sqrt(14.477^2 - 4*5.7*4.9)) / 9.8
We will use the "plus" because the "minus" was on the way UP.
t = 2.49 seconds
Total x = (10.91 m/s) (2.49 s) = 27.17 m
This is 24 + 3.17, so the distance from the outer surface of the wall
to the point where the ball lands on the roof is 3.17 meters.
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