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.
a) Find the speed at which the ball was launched.m/s
b) Find the vertical distance by which the ball clears the wall. m
c Find the horizontal distance from the wall to the point on the roof where the ball lands. m
I posted this question earlier and I didn't understand the feedback. Could someone explain it differently? I need it step by step to understand it. Thanks!
I provided the "feedback", in steps. What didn't you understand about it?
For the speed at which the ball was launched I got 18.12m/s. That part was right according to the computer. I didn't understand parts b or c of the question. For part a you multiplied vt by .608 instead of .6018 and that is why I got a different answer. Could you explain parts b and c again? Thanks.
a) 18.12
Sure, I'd be happy to help explain the problem step by step!
Given:
- The height of the school building, h = 6.50 m
- The height of the railing, r = 1.3 m
- The horizontal distance from the base of the wall to the point where the ball is launched, d = 24.0 m
- The angle at which the ball is launched above the horizontal, θ = 53.0°
- The time taken by the ball to reach a point vertically above the wall, t = 2.20 s
To solve the problem, we'll use the following principles of projectile motion:
1. The horizontal and vertical motion of a projectile can be treated independently.
2. The horizontal motion is at a constant velocity, while the vertical motion is accelerated due to gravity.
Now let's break down the problem into smaller steps:
Step 1: Find the initial horizontal velocity and horizontal distance covered
- The horizontal component of velocity remains constant throughout the motion.
- We know the horizontal distance, d = 24.0 m.
- The equation for horizontal distance is given by: d = (v₀x) * t, where v₀x is the initial horizontal velocity (which is constant throughout the motion).
- Rearranging the equation, we can solve for v₀x: v₀x = d / t.
Step 2: Find the initial vertical velocity
- The vertical component of velocity changes due to acceleration caused by gravity.
- At the highest point of its trajectory (when the ball is vertically above the wall), the vertical velocity becomes zero (v_y = 0).
- We can use the kinematic equation for vertical velocity: v_y = v₀y + a_y * t
- Since the acceleration in the vertical direction is due to gravity, a_y is -9.8 m/s² (negative because it acts downward).
- Rearranging the equation, we can solve for v₀y: v₀y = -a_y * t.
Step 3: Find the initial launch velocity and horizontal component of velocity
- The initial launch velocity, v₀, can be found using the following equation: v₀ = √(v₀x² + v₀y²)
- Substitute the values we found for v₀x and v₀y to calculate the initial launch velocity.
Step 4: Find the vertical distance cleared
- At the highest point of its trajectory, the vertical displacement is equal to the height of the wall plus the height of the railing (6.50 m + 1.3 m).
- The equation for vertical displacement is: Δy = v₀y * t + (1/2) * a_y * t².
- Substitute the known values of v₀y, t, and a_y to calculate the vertical distance cleared.
Step 5: Find the horizontal distance to the landing point
- The total time of flight, T, can be calculated using the equation: T = 2 * t (since the ball takes the same time to reach its highest point and descend back down).
- The horizontal distance covered during the total time of flight is given by: d_total = v₀x * T.
- Substitute the known values of v₀x and T to calculate the horizontal distance to the landing point.
Now, use these steps to calculate the answers to the given questions:
a) Find the speed at which the ball was launched.
- Follow Step 3 to calculate the initial launch velocity.
b) Find the vertical distance by which the ball clears the wall.
- Follow Step 4 to calculate the vertical distance cleared.
c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
- Follow Step 5 to calculate the horizontal distance to the landing point.
I hope this step-by-step explanation helps you understand the problem better! Let me know if you have any further questions.