Question

A playground is on the flat roof of a city school, 5.9 m above the street below (see figure). The vertical wall of the building is h = 7.40 m high, forming a 1.5-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 θ 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. (Give your answer to two decimal places to reduce rounding errors in later parts.) m/s (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

Answer 1

(a) Here, horizontal distance x = 24 = u*2.2 cos 53
=> u = 18.13 m/s

So, the speed at which the ball was launched = 18.13 m/s

(b) The vertical distance travelled in this time is ut – ½ a t^2

y = 18.13 *2.2 sin 53 – (1/2)*9.8*2.2^2 = 8.14 m

The distance above the wall is 8.14 – 7.40 = 0.74 m

(c) Given that the vertical distance of the play ground from the ground is 5.9 m
The time to fall on the play ground is found from
5.9 = 18.13 *t* sin 53 – (1/2)*9.8*t^2

=> 5.9 = 14.5t - 4.9t^2

=> t^2 - 3t + 1.2 = 0

=> t = [3 + sqrt(9 - 4.8)] / 2 = [3 + 2.05]/2 = 2.52 s

other value of t is neglected since it is less than 2.20 s

The horizontal distance traveled in this time is
18.13 *2.52cos 53 = 27.50 m