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A playground is on the flat roof of a city school, 6.3 m above the street below (see figure). The vertical wall of the building is h = 7.40 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

Edit
Added Thu, 21 Apr '16
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Horizontal component of ball's velocity is

24 m/2.20 s = 10.91 m/s

So the ball's initial speed was

10.91 m/s divided by cos(53 degrees) = 18.127 m/s

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 7.4-meter wall by 0.73 m

ball lands when y = 6.3 m on the way down.

5.3 = 14.477 t - (4.9) t^2

0 = 4.9t^2 - 14.477 t + 6.3

t = (14.477 plus or minus sqrt(14.477^2 - 4*6.3*4.9)) / 9.8

We will use the "plus" because the "minus" was on the way UP.

t = (14.477 plus sqrt(85.9)) /9.8

= 2.42 seconds

Total x = (10.91 m/s) (2.42 s) = 26.40m

This is 24 + 2.4, so the distance from the outer surface of the wall

to the point where the ball lands on the roof is 2.4 meters.

Edit
Added Thu, 21 Apr '16
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