A playground is on the flat roof of a city school, 5.7 m above the street below (see figure). The vertical wall of the building is h = 7.10 m high, to form a 1.4-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.
Find the speed at which the ball was launched.
m/s
Find the vertical distance by which the ball clears the wall.
m
Find the horizontal distance from the wall to the point on the roof where the ball lands.
m
Well, it seems like the ball's journey is quite an adventure! Let me help you with your calculations, but in my own humorous way!
To find the speed at which the ball was launched, we can start by breaking down the ball's motion into its horizontal and vertical components. The time it takes for the ball to reach vertically above the wall can give us the vertical component of velocity. So, let's call that vy.
Using the equation for vertical displacement (y), we can say that y = vy * t - 0.5 * g * t^2, where g is the acceleration due to gravity. Since the ball started at a height of 5.7 m and reached a point vertically above the wall, the vertical displacement is h - 1.4 m.
So, h - 1.4 = vy * 2.2 - 0.5 * 9.8 * 2.2^2. Solving this equation will give us the value of vy, which we can then use to find the initial speed.
Now, for the vertical distance by which the ball clears the wall, we need to find the maximum height the ball reaches above the roof. We can use the equation for vertical displacement again, but this time, the initial vertical velocity will be vy, and the final vertical velocity will be 0 m/s.
So, using the equation y = vy^2 / (2 * g), we can solve for y. This distance will give us the vertical distance by which the ball clears the wall.
Lastly, to find the horizontal distance from the wall to the point on the roof where the ball lands, we can use the horizontal component of velocity, which we'll call vx. The time it takes for the ball to reach vertically above the wall will also give us the time it takes for it to land on the roof.
Using the equation x = vx * t, where x is the horizontal distance, we can solve for x.
So, these are the steps you need to take to find the answers. Plug in the values and solve those equations!
To solve this problem, we can break down the motion of the ball into its horizontal and vertical components.
First, let's find the initial vertical velocity of the ball:
Using the equation of motion for vertical motion:
y = yi + viy*t - (1/2)*g*t^2
Since the ball starts at a point vertically above the wall, the initial vertical position (yi) is the height of the railing (1.4 m).
Plugging in the values:
1.4 m = 0 + viy*2.20 s - (1/2)*9.8 m/s^2*(2.20 s)^2
Solving for viy:
viy = (1.4 m + (1/2)*9.8 m/s^2*(2.20 s)^2)/(2.20 s)
viy = 3.235 m/s
Next, let's find the initial horizontal velocity of the ball:
Using the equation for horizontal velocity:
vix = d/t
Plugging in the values:
vix = 24.0 m/2.20 s
vix = 10.909 m/s
Now we can find the initial velocity of the ball:
Using the equation for initial velocity:
vi = √(vix^2 + viy^2)
Plugging in the values:
vi = √(10.909 m/s)^2 + (3.235 m/s)^2
vi = √118.9 m^2/s^2
vi = 10.913 m/s
So, the speed at which the ball was launched is approximately 10.913 m/s.
Next, let's find the vertical distance by which the ball clears the wall:
Using the equation of motion for vertical motion:
y = yi + viy*t - (1/2)*g*t^2
Since the ball reaches a point vertically above the wall, the final vertical position is 7.1 m (height of the wall).
Plugging in the values:
7.1 m = 1.4 m + 3.235 m/s*2.20 s - (1/2)*9.8 m/s^2*(2.20 s)^2
Solving for y:
y = 7.1 m - 1.4 m - 7.1054 m
y ≈ -1.4054 m
Since the result is negative, it means the ball clears the wall by approximately 1.4054 m above it.
Finally, let's find the horizontal distance from the wall to the point on the roof where the ball lands:
Using the equation for horizontal distance:
x = vix*t
Plugging in the values:
x = 10.909 m/s*2.20 s
x ≈ 24.0 m
So, the horizontal distance from the wall to the point on the roof where the ball lands is approximately 24.0 m.
To find the speed at which the ball was launched, we can use the formula for horizontal range of a projectile.
The horizontal range is the distance traveled by the ball in the horizontal direction before hitting the ground. We can calculate the horizontal range using the equation:
Range = (Initial Velocity * Time of Flight) * Cos(theta),
where "Initial Velocity" is the speed at which the ball was launched, "Time of Flight" is the time taken by the ball to reach a point vertically above the wall, and "Cos(theta)" is the cosine of the launch angle.
Given that the range is equal to the horizontal distance from the wall, which is 24.0 m:
24.0 m = (Initial Velocity * 2.20 s) * Cos(53.0°).
Now, we can solve this equation to find the value of "Initial Velocity".
Next, to find the vertical distance by which the ball clears the wall, we need to calculate the maximum height reached by the ball above the level of the roof. We can use the formula for maximum height of a projectile.
The maximum height is given by:
Max height = (Initial Velocity * Sin(theta))^2 / (2 * g),
where "g" is the acceleration due to gravity.
Given that the maximum height is equal to the height of the building plus the height of the railing:
Max height = (7.10 m + 1.4 m).
Now, we can solve this equation to find the value of "Initial Velocity".
Finally, to find the horizontal distance from the wall to the point on the roof where the ball lands, we can use the equation of horizontal range mentioned earlier.
Now, let's solve the equations to find the answers.