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 = 6.50 m high, forming a 1.3-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 è = 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.
Help please! Thanks.:)
I don't see a question here.
(a) Find the speed at which the ball was launched.
(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.
To solve this problem, we can break it down into several steps:
Step 1: Calculate the initial velocity components of the ball.
We can start by finding the initial velocity components of the ball. The horizontal component of the velocity (Vx) can be found using the formula Vx = V * cos(θ), where V is the initial velocity and θ is the launch angle. Similarly, the vertical component of the velocity (Vy) can be found using Vy = V * sin(θ).
Step 2: Calculate the time it takes for the ball to reach the top of the trajectory.
Since the ball takes 2.20 s to reach a point vertically above the wall, we can calculate the time it takes for the ball to reach the top of its trajectory. This time can be found using the formula t = Vy / g, where g is the acceleration due to gravity.
Step 3: Calculate the maximum height of the ball above the playground.
Using the time calculated in step 2, we can find the maximum height (H) reached by the ball above the playground. This can be found using the formula H = Vy * t - 0.5 * g * t^2.
Step 4: Calculate the horizontal distance traveled by the ball.
Next, we need to find the horizontal distance (D) traveled by the ball. This can be calculated using the formula D = Vx * t, where Vx is the horizontal component of the initial velocity and t is the time calculated in step 2.
Step 5: Determine if the ball clears the wall.
Finally, we can check if the ball clears the wall by comparing the maximum height (H) calculated in step 3 with the height of the wall (h + 1.3 m). If the maximum height is greater than the wall height, then the ball clears the wall. Otherwise, it does not.
By following these steps and substituting the given values into the formulas, you should be able to find the answers you need.
To solve this problem, we will break it down into several steps:
Step 1: Determine the initial vertical velocity of the ball.
Since the ball takes 2.20 s to reach a point vertically above the wall, we can use the kinematic equation:
h = v0t + (1/2)gt^2
Given:
h = 6.5 m
t = 2.20 s
g = acceleration due to gravity = 9.8 m/s^2
By substituting the values into the equation, we can solve for the initial vertical velocity, v0.
Step 2: Determine the horizontal velocity of the ball.
The horizontal velocity of the ball remains constant throughout its motion. We can use the horizontal distance traveled, d = 24.0 m, and the time taken, t = 2.20 s, to calculate the horizontal velocity, v_h.
Using the equation:
d = v_ht
Rearranging the equation gives:
v_h = d / t
Step 3: Calculate the initial velocity and angle of projection.
Once we have the initial vertical velocity, v0, and the horizontal velocity, v_h, we can determine the magnitude of the initial velocity, v, and the angle of projection, ø.
The initial velocity, v, can be calculated using the Pythagorean theorem:
v = sqrt(v_h^2 + v0^2)
The angle of projection, ø, can be calculated using the formula:
ø = arctan(v0 / v_h)
Step 4: Calculate the time of flight.
The time of flight, T, is the total time the ball remains in the air. We can use either the vertical or horizontal motion to calculate it.
Using the horizontal distance, d = 24.0 m, and the horizontal velocity, v_h, we can calculate the time of flight using the formula:
T = d / v_h
Step 5: Calculate the maximum height reached by the ball.
The ball reaches its maximum height at the halfway point of its vertical motion. The time taken to reach the maximum height is half the time of flight, T.
Using the formula:
h_max = v0 * (T/2) - (1/2)g * (T/2)^2
Step 6: Determine if the ball clears the railing.
The maximum height reached by the ball, h_max, should be greater than the height of the railing. If it is, then the ball clears the railing. Otherwise, it hits the railing.
By following these steps and plugging in the given values, you should be able to find the answers you need.