A playground is on the flat roof of a city school, 7.00 m above the street below. The vertical wall of the building is 8.00 m high, forming a 1.00 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 53.0° above the horizontal at a point 24.0 m from the base of the building wall. The ball takes 2.00 s to reach a point vertically above the wall.
I had solved for A, but I am unable to solve for B and C, it says that I was off by 10-100%.
(a) Find the speed at which the ball was launched. 19.9 m/s
(b) Find the vertical distance by which the ball clears the wall.
(c) Find the distance from the wall to the point on the roof where the ball lands.
Assuming a) is right.
b) how long does it take to get to the wall plane? time= 24/(19.9cosTheta)
Now using that time..
vertical height= 19.9sinTheta*t-1/2 g t^2
and clearance is that minus 8m
To solve for part (b) and part (c), we need to find the vertical and horizontal components of the ball's initial velocity. We already know the launch angle (θ) is 53.0° and the time it takes to reach a point vertically above the wall (t) is 2.00 s.
Let's start by finding the vertical component of the initial velocity (Vy). We can use the equation of motion:
Vy = (Δy) / t
The vertical displacement (Δy) is the height of the wall plus the height of the railing: Δy = 8.00 m + 1.00 m = 9.00 m
Vy = 9.00 m / 2.00 s = 4.50 m/s
Now, let's find the horizontal component of the initial velocity (Vx) using the following equation:
Vx = (Δx) / t
The horizontal displacement (Δx) is given as 24.0 m.
Vx = 24.0 m / 2.00 s = 12.0 m/s
Now we can move on to solving part (b) and part (c).
(b) Find the vertical distance by which the ball clears the wall:
To find this distance, we can use the equation of motion for vertical motion:
Δy = Vy * t + (1/2) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the known values:
Δy = (4.50 m/s) * 2.00 s + (1/2) * (9.8 m/s^2) * (2.00 s)^2
Δy = 9.00 m + 19.6 m
Δy = 28.6 m
Therefore, the ball clears the wall by 28.6 m vertically.
(c) Find the distance from the wall to the point on the roof where the ball lands:
To find this distance, we can use the equation of motion for horizontal motion:
Δx = Vx * t
Substituting the known values:
Δx = (12.0 m/s) * 2.00 s
Δx = 24.0 m
Therefore, the distance from the wall to the point on the roof where the ball lands is 24.0 m.
To solve parts (b) and (c) of the problem, we need to use the kinematic equations of projectile motion. Let's break down the problem and find the solutions step by step.
Step 1: Determine the initial vertical and horizontal velocity components of the ball.
Given:
- Angle of launch (θ) = 53.0°
- Distance from the base of the building wall (x) = 24.0 m
First, we need to find the initial vertical velocity (Vy) and initial horizontal velocity (Vx) of the ball.
Using trigonometry:
Vy = V_initial * sin(θ)
Vx = V_initial * cos(θ)
Step 2: Find the time it takes for the ball to reach the point vertically above the wall (2.00 s).
Given:
- Time of flight (t) = 2.00 s
Step 3: Calculate the vertical distance cleared by the ball (Δy).
Using the equation of motion:
Δy = V_initial_y * t - (1/2) * g * t^2
We know that the initial vertical velocity (V_initial_y) is Vy and the acceleration due to gravity (g) is approximately 9.8 m/s^2.
Step 4: Calculate the distance from the wall to the point on the roof where the ball lands (D).
Since the horizontal velocity remains constant throughout the motion, we can use the equation:
D = V_initial_x * t
Step 1: Find the initial vertical and horizontal velocity components:
Vy = V_initial * sin(53.0°)
Vx = V_initial * cos(53.0°)
Step 2: Find the time it takes for the ball to reach the point vertically above the wall:
t = 2.00 s
Step 3: Calculate the vertical distance cleared by the ball:
Δy = Vy * t - (1/2) * g * t^2
Step 4: Calculate the distance from the wall to the point on the roof where the ball lands:
D = Vx * t
Now that we have the steps, let's solve for parts (b) and (c) using the given information.
(b) Find the vertical distance by which the ball clears the wall.
Substituting the known values into the equation for Δy:
Δy = Vy * t - (1/2) * g * t^2
Plug in the values:
V_initial = 19.9 m/s (from part a)
θ = 53.0°
t = 2.00 s
g = 9.8 m/s^2 (acceleration due to gravity)
Vy = V_initial * sin(θ)
Δy = Vy * t - (1/2) * g * t^2
Substituting the known values:
Vy = 19.9 m/s * sin(53.0°)
Δy = (19.9 m/s * sin(53.0°)) * 2.00 s - (1/2) * 9.8 m/s^2 * (2.00 s)^2
Simplify the equation and solve for Δy.
(c) Find the distance from the wall to the point on the roof where the ball lands.
Using the equation for D:
D = V_initial_x * t
Plug in the known values:
V_initial = 19.9 m/s (from part a)
θ = 53.0°
t = 2.00 s
Vx = V_initial * cos(θ)
D = V_initial * cos(θ) * t
Simplify the equation and solve for D.
By following these steps and plugging in the given values, you should be able to find the solutions for parts (b) and (c) of the problem.