Aplayground is on the flat roof of a city school 6.2 m above the street bek (see

Question

Aplayground is on the flat roof of a city school 6.2 m above the street bek (see figure). The vertical sul of the bu ding ishs170m high, forming a 1.5 ighraling around the playground. A bal has fallen to the street below and a passerby returns it by launching it at an angle of··53.or 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 wal c) Find the horizontal distance from the wall to the point on the roof where the ball lands Need Help? Modn 02 6 8

Explanation / Answer

Apply kinematic equation along horizontal direction

xf= xi + vxi t + 1/2 ax t^2

24 m = 0 + vi cos 53( 2.2) + 0 ( ax =0)

vi = 18.1 m/s

(b)

along y direction

yf = yi + vyi t + 1/2 ay t^2

=0 + 18.12 sin 53 ( 2.2 s) + 1/2 * ( -9.8 m/s^2) ( 2.2 s)^2

= 8.13 m

so it clears the parapet by 8.13 m - 7.7 = 0.43 m

(c)

Apply kineatic equation

yi = ( tan theta) xi - ( g/ 2 vi^2 cos^2 theta) xi^2

6.2= tan ( 53) xi - ( 9.8/2(18.1) cos^2 53) xi^2

0.0412 xi^2 -1.33 xi +6.2 m= 0

solving quadratic equation

xi = 26.63 m or 5.65 m

The ball passes twice through the level of the roof.
It hits the roof at distance from the wall

26.63 -24 m = 2.63 m

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