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.
I got 1.233104959 m for b. I just need help with c.
To find the answer to part (b), the vertical distance by which the ball clears the wall, we can use the equation for vertical displacement:
𝑑𝑦 = 𝑣𝑖𝑦𝑡 + 0.5𝑎𝑡^2
Since the ball is launched at an angle, we need to split the initial velocity into its vertical and horizontal components. The vertical component of the initial velocity, 𝑣𝑖𝑦, can be found using:
𝑣𝑖𝑦 = 𝑣𝑖 * sin(𝜃)
where 𝜃 is the launch angle. Given that 𝜃 = 53.0° and the speed of the ball is 𝑣𝑖 = 18.127 m/s (found in part (a)), we can calculate 𝑣𝑖𝑦:
𝑣𝑖𝑦 = 18.127 * sin(53.0°)
Next, we need to find the acceleration, 𝑎. In this case, the only force acting on the ball in the vertical direction is gravity, so the acceleration is given by 𝑔, the acceleration due to gravity. 𝑔 is approximately 9.8 m/s^2.
Finally, we can find the time, 𝑡, it takes for the ball to reach a point vertically above the wall. The given information tells us that 𝑡 = 2.20 s.
Now we can substitute the known values into the equation for vertical displacement to find the answer to part (b):
𝑑𝑦 = 𝑣𝑖𝑦𝑡 + 0.5𝑎𝑡^2
Substituting the values:
𝑑𝑦 = (18.127 * sin(53.0°)) * (2.20) + 0.5 * (9.8) * (2.20^2)
Calculating this expression will give us the vertical distance by which the ball clears the wall.
To find the answer to part (c), the horizontal distance from the wall to the point on the roof where the ball lands, we can use the equation for horizontal displacement:
𝑑𝑥 = 𝑣𝑖𝑥𝑡
The horizontal component of the initial velocity, 𝑣𝑖𝑥, can be found using:
𝑣𝑖𝑥 = 𝑣𝑖 * cos(𝜃)
Substituting the known values:
𝑑𝑥 = (18.127 * cos(53.0°)) * (2.20)
Calculating this expression will give us the horizontal distance from the wall to the point where the ball lands.
After performing the necessary calculations, we can obtain the values for parts (b) and (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.