A playground is on the flat roof of a city school, 6.00 m above the street below

Question

A playground 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 2.00 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 distance from the wall to the point on the roof where the ball lands.
m

Explanation / Answer

This really isnt that difficult... I use this one as simple example for my students, before I assign them the harder problems.

You have to do part a first, to determine the speed. To do so, write the equation

    final position = initial position + initial velocity * time + (1/2) a t^2

for the horiz direction

    26 = 0 + v cos53 * 2.00 + 0

solve for v

    v = 26 / 2.00 cos53 = 21.601 m/s

Now for part (c). Recognize that when the ball hits the roof, its height is 6 m above the ground. So the final vertical position is 6. You just have to find the final horiz position.

horizontal equation     x = 0 + 21.601*cos53*t + 0   (note that time is now unknown)

vertical equation       6 = 0 + 21.601*sin53*t - (1/2)*9.8*t^2

Notice that t is the only unknown in the second equation. It is a quadratic equation with:

     0 = 4.9 t^2 - 17.252 t + 6

Using quadratic formula, we get t = 3.13 seconds

Now calculate x

   x = 21.601 * cos53 * 3.13 = 40.68 meters

This is the distance from the man, so the distance the ball lands from the wall is

   40.68 - 26 =   14.68 meters

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