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.

Question

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
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

(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.

Answer 1

Check part b comments on your previous post.

Answer 2

thanks for your help!

Answer 3

For b.)

You got the y component correct however you multiplied it by sin53 a second time.
The correct equation should be y=18.13sin53(2.2)-4.9(4.8)

or with the y component solved for already, y=14.48(2.2)-4.9(4.8)

The ball is at height= 8.336m
It clears the wall by 8.336-6.7=1.636m

Answer 4

For c.)

Set the y component to 5.2m

5.2= 18.13sin53(t)-4.9(t)^2
solve the quadratic for t and plug into d=18.13cos53(t)

This will give you the horizontal distance until the ball lands on the roof.

Answer 5

To solve these problems, let's break them down step by step.

(a) Finding the speed at which the ball clears the wall:
To find the speed at which the ball clears the wall, we need to analyze the horizontal and vertical components of the ball's velocity separately.

The horizontal component of the initial velocity (v₀x) remains constant throughout the motion, so we can use the equation: v₀x = v₀ * cos(θ)

Given that θ = 53.0° and you found the correct value for v₀ (18.13 m/s), you can calculate v₀x: v₀x = 18.13 * cos(53.0°)

(b) Finding the vertical distance by which the ball clears the wall:
In this case, we need to calculate the vertical displacement of the ball. It is important to note that the ball is launched from a certain height (h), and it takes 2.20 s to reach a point vertically above the wall (so it is in the air for 2.20 s).

To find the vertical displacement (dy), we can use the equation: dy = v₀y * t + (1/2) * g * t²

In this equation, v₀y represents the initial vertical velocity, t is the time of flight, and g is the acceleration due to gravity (-9.8 m/s²).

You correctly found v₀y = v₀ * sin(θ) = 18.13 * sin(53.0°). Now, substitute the values into the equation to calculate dy.

(c) Finding the horizontal distance from the wall to the point on the roof where the ball lands:
To find the horizontal distance, we can use the equation: dx = v₀x * t

You already have the value of v₀x from part (a), and the time of flight (t) is given as 2.20 s. Plug these values into the equation to get the horizontal distance.

By following these steps, you should be able to find the answers to parts (b) and (c) of the question.