In: Physics
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 28.0 m from the base of the building wall. The ball takes 2.10 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
Part (a): The horizontal distance is x = 28.0 m, the launching angle is 53.0o and the time taken to reach that distance is t = 2.10 s. Thus the initial velocity can be calculated as,
Part (b): The vertical displacement is,
Thus the ball clears the wall by distance of,
Part (c): According to the path equation of motion of ball,
So the distance from the wall is,