A playground is on the flat roof of a city school, 5.1 m above the street below

Question

A playground is on the flat roof of a city school, 5.1 m above the street below (see figure). The vertical wall of the building is h = 6.50 m high, forming a 1.4-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 orizontal at a point 2.0m 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.

Explanation / Answer

A) Let launch speed be v.

v cos 53 degree *t = d

v = d/( t cos 53 degree)

= 24/(2.2* cos 53 degree)

= 18.13 m/s answer

B) vertical distance by which it clears the wall

= height of the ball - h

= v sin 53 degree *t - 0.5gt^2 - h

= 18.13*sin 53 degree *2.2 - 4.9*2.2^2 - 6.5

= 1.64 m answer

C) Lets find the landing time first,

5.1 = 18.13*sin 53 degree *t - 4.9*t^2

taking the larger t, t =2.546s

Distance by which it clears the wall

= 18.13 cos 53 degree *2.546 - 24

= 3.78 m answer

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