A ball has fallen from a playground on a
at roof of a city school building that is 10:0m
above the street level.The vertical wall of the building is 11m high, forming a 1m railing
around the playground. A passerby returns the ball by launching it at an angle of 53
degrees above the horizontal at a point 25m from the base of the building wall. The ball
takes 2:0 s to rich a point vertically above the wall.
(a) Determine the speed with which the ball was launched.(4 Points.)
(b) Find the vertical distance by which the ball clears the wall. (4 Points)

Question

A ball has fallen from a playground on a 
at roof of a city school building that is 10:0m
above the street level.The vertical wall of the building is 11m high, forming a 1m railing
around the playground. A passerby returns the ball by launching it at an angle of 53
degrees above the horizontal at a point 25m from the base of the building wall. The ball
takes 2:0 s to rich a point vertically above the wall.
(a) Determine the speed with which the ball was launched.(4 Points.)
(b) Find the vertical distance by which the ball clears the wall. (4 Points)

anonymous · Answer

(c) Find the distance from the wall to the point on the roof where the ball lands.(6
Points.)
(d) Determine the horizontal and vertical components of the velocity of the ball at the
time of landing.(6 Points.)

anonymous · Answer

Alright, I'm trying to draw a picture of this to see what's going on...

anonymous · Answer

Ok, here's the main thing that's going on:

There is a parabolic trajectory (neglecting air resistance) defined by:

$x=v_xt$
$y=v_yt-0.5gt^2$

anonymous · Answer

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anonymous · Answer

"The ball takes 2:0 s to rich a point vertically above the wall." means that in $x=v_xt$, x=25m, t=2.0; you can solve for $v_x$

anonymous · Answer

i solved the entire thing i got 
a) = 20.77
b) = 2.58
c) = 32.5
d) = -8.89

anonymous · Answer

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