A playground is on the flat roof of a city school, 6.4 m above the street below

Question

A playground is on the flat roof of a city school, 6.4 m above the street below (see figure). The vertical wall of the building is h = 7.90 m high, to form 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 = 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. m/s (b) Find the vertical distance by which the ball clears the wall. m (c) Find the horizontal distance from the wall to the point on the roof where the ball lands. m

Explanation / Answer

a) suppose intial velocity is u then

horizontal vleocity. ux = u cos53

it travels 24 m in 2.20 s horizontally.

24 = ( u cos53) (2.20)

u = 18.13 m/s

b) vertical height at this time,

y= uy * t + ay ^ t^2 /2

y = (18.13 sin53 2.20) + (-9.81 x 2.2^2 / 2)

y = 8.11m

height at which ball clears wall, h = 8.11 - 7.90 = 0.21 m

c) ball landa at height 6.4m from ground.

6.4 = (18.13sin53)t - 9.81t^2 /2

4.905 t^2 - 14.48t + 6.4 = 0

t = 2.50 sec

horizontal distance traveled by ball = ux * t
= 18.13 cos53 * 2.50 = 27.28 m

distane form wall = 27.28 - 24 = 3.28 m

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