A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). The vertical wall of the building is h = 6.70 m high, forming 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 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 clears the wall.
I calculated this correctly and got 18.13 m/s.
(b) Find the vertical distance by which the ball clears the wall.
I know that dy=vy0*t but I tried using that equation and couldn't get the answer.
I did this and it was incorrect
18.13*sin53*2.2=dy
32.85-6.7=25.45
(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
I can't figure out how to do the last two, can anyone please help me set up these problems so I can figure out how to solve them? Thank you.
On part b,
You need to find how long it takes the ball to go th 24.0m in the x direction. Use that time for the y calculation.
Sorry, I see that was given.
Yeah :( thanks for trying though, I tried using the v0sin53 and got 18.13*sin53= 14.48
then I tried plugging that into the y, to get y=14.48sin53(2.2)-1/2(-9.8)(2.2^2)
I got 25.44-23.716=1.725, the computer assignment says that is incorrect. I don't know what I am doing wrong
I calculated the ball was thrown at a speed of 18.12m/s.
Vx is 24m/2.2s=10.91m/s
Vx=V(cos(53))
That gives V as 18.12m/s
I get 14.47m/s for Vy.
Using that...
Y=14.47m/s(2.2s)-(1/2)(9.8m/s)(2.2)^2
Y=31.834m - 23.716m
=8.118m
So it clears the wall by 8.118m - 6.7m
=1.418m
To find the vertical distance by which the ball clears the wall (part b), you can use the equation for vertical displacement:
dy = vy0 * t + (1/2) * g * t^2
Where:
- dy is the vertical distance traveled by the ball
- vy0 is the initial vertical velocity of the ball
- t is the time it takes for the ball to reach a point vertically above the wall
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
We already know the value of t (2.2 s) and we can calculate vy0 using the initial launch angle and speed.
Given:
- Launch angle θ = 53.0°
- Speed of the ball = 18.13 m/s (which you calculated correctly)
To calculate the initial vertical velocity (vy0), you can use the equation:
vy0 = v * sin(θ)
Where:
- v is the speed of the ball
- θ is the launch angle in radians
First, convert the angle from degrees to radians:
θ_radians = θ * (π/180) = 53.0° * (π/180) = 0.925 radians (rounded to three decimal places)
Now, calculate vy0:
vy0 = 18.13 m/s * sin(0.925 radians) ≈ 16.67 m/s (rounded to two decimal places)
Now, substitute the values into the initial equation and solve for dy:
dy = (16.67 m/s) * (2.2 s) + (1/2) * (9.8 m/s^2) * (2.2 s)^2
dy = 36.674 m + 23.7164 m
dy ≈ 60.39 m
So, the ball clears the wall by approximately 60.39 meters vertically.
To find the horizontal distance from the wall to the point on the roof where the ball lands (part c), we need to use the equation for horizontal displacement:
dx = vx * t
Where:
- dx is the horizontal distance traveled by the ball
- vx is the initial horizontal velocity of the ball
In this case, the horizontal velocity (vx) remains constant throughout the trajectory because there is no horizontal acceleration.
To calculate vx, you can use the equation:
vx = v * cos(θ)
Where:
- v is the speed of the ball
- θ is the launch angle in radians
Again, use the value of θ in radians that we calculated earlier (0.925 radians):
vx = 18.13 m/s * cos(0.925 radians) ≈ 5.19 m/s (rounded to two decimal places)
Now, substitute the values into the equation for dx:
dx = (5.19 m/s) * (2.2 s)
dx ≈ 11.42 m (rounded to two decimal places)
So, the ball lands approximately 11.42 meters from the base of the building wall on the roof of the school playground.