A playground is on the flat roof of a city school, hb = 5.70 m above the street below (see figure). The vertical wall of the building is h = 6.90 m high, to form a 1.2-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.
18.127
Correct: Your answer is correct.
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.
2 answers
I got 1.233104959 m for b. I just need help with c.
well for part c it still has the same old horizontal velocity
u = 18.1 cos 53
so
x = u t where t is the time to fall from the point where it it is above the wall up there to the roof.
call v the vertical velocity up when it is directly above the wall and 1.23 meters above it
then we have
h = wall height + height above wall + v t - 4.9 t^2
when h = 0, we are at the roof
solve that for t
use the longer t of the two results, the first solution to the quadratic was at the roof as the ball went up.
u = 18.1 cos 53
so
x = u t where t is the time to fall from the point where it it is above the wall up there to the roof.
call v the vertical velocity up when it is directly above the wall and 1.23 meters above it
then we have
h = wall height + height above wall + v t - 4.9 t^2
when h = 0, we are at the roof
solve that for t
use the longer t of the two results, the first solution to the quadratic was at the roof as the ball went up.
1 answer