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 h = 7.0 m high, to form a 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 ? = 53.00 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.

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

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

lands.

Explanation / Answer

Let the velocity of the ball be v
The vleocity of the ball can be expressed as vcos53i + vsin53j.

Now the horizontal distance between the passerby and the wall is 24
hence
vcos53 = 24/2.20
v = 24/(2.2*cos53)
v = 18.13 m/s

a) v = 18.13 m/s
b) Initial Vertical velocity = vsin53

s = ut - 1/2*a*t^2*a*t^2
s = 18.13*sin53*2.2 -1/2*9.8*2.2^2
s = 8.14 m
c)
So for that we need to find the time taken
Hence
s = ut + 1/2*a*t^2
6 = 18.13*sin53*t -1/2*9.8*t^2
Solving the quadratic we get,
t = 0.498474, t = 2.45648

Hence the time would be 2.45648

Total Horizontal distance travelled in the time 2.45648 sec is :
s = 18.13*cos53*2.45648
s = 26.8024 m

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