A playground is on the flat roof of a city school, 5.8 m above the street below (see figure). The vertical wall of the building is h = 7.30 m high, to form 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 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

a) The horizontal component of the launch speed (Vo) is Vo cos = 0.6018 V. It travels 24 m horizontally in 2.20 s. Therefore

X= X_0+V_x0(t)
24= 0+V_x(2.2)
10.91= V_0cos53
V_0= 18.13

b)to solve for the height after t = 2.2s
y = (Vo sin 53)* t - (1/2) g t^2
= ((18.13sin53)*2.2)- 4.9(2.2^2)
= 31.86-23.72
y = 8.14
8.14-7.3= .84

c) Use the same equation to solve for t when y = 5.8 m
5.8= ((18.13sin53)t)-4.9t^2
= 4.9t^2+14.18t+5.8
I suggest that you graph this on your calculator to solve for t
whatever your answer is for t plug it in to this equation:
x= V_x(t)
then subtract 24.0 m from your answer which will finally give u the answer for c

Oh, I see. You want me to solve a physics problem for you. Well, I'm not exactly the most inclined when it comes to math, but I'll give it a shot. Brace yourself for some math humor!

(a) To find the speed at which the ball was launched, we can use the formula for projectile motion. But since I prefer projectile jokes over physics, I'll launch into a pun instead: Why don't scientists trust atoms? Because they make up everything!

(b) Now, let's calculate the vertical distance by which the ball clears the wall. I must say, this ball is quite the high flier! It's like the Michael Jordan of projectiles. Anyway, to get the answer, we can use the basic kinematic equation. I guess you could say physics problems are pretty elementary, my dear Watson.

(c) Lastly, let's determine the horizontal distance from the wall to the point on the roof where the ball lands. This should be easy peasy, lemon squeezy! Just another quick calculation using the same kinematic equation. Talking about calculations, did you hear about the mathematician who's afraid of negative numbers? He'll stop at nothing to avoid them!

Alright, I hope my clown-like attempt at answering your questions was at least mildly amusing. If not, just remember that laughter is the best way to tackle any problem.

To solve this problem, we can break it down into different components and use the equations of motion to find the answers.

(a) Find the speed at which the ball was launched:
To find the speed, we can use the horizontal component of the initial velocity since the vertical component is not given. We can use the equation:

v = d / t

where v is the velocity, d is the horizontal distance, and t is the time taken.

Given: d = 24.0 m and t = 2.20 s
Plugging in these values, we get:

v = 24.0 m / 2.20 s
v ≈ 10.91 m/s

Therefore, the speed at which the ball was launched is approximately 10.91 m/s.

(b) Find the vertical distance by which the ball clears the wall:
To determine the vertical distance the ball clears the wall, we need to find the vertical component of the projectile's velocity. We can use the equation:

y = y0 + v0y * t + (1/2) * a * t^2

where y is the vertical distance, y0 is the initial vertical position, v0y is the initial vertical velocity, t is the time taken, and a is the acceleration (in this case, due to gravity).

Given: y0 = 5.8 m, t = 2.20 s, and a = -9.8 m/s^2 (acceleration due to gravity)
To find v0y, the initial vertical velocity, we can use the equation:

v0y = v * sin(θ)

where v is the initial speed and θ is the launch angle.

v = 10.91 m/s (as found in part (a))
θ = 53.0°

Plugging in these values, we get:

v0y = 10.91 m/s * sin(53.0°)
v0y ≈ 8.71 m/s

Now we can find the vertical distance by plugging the values into the equation:

y = 5.8 m + (8.71 m/s) * (2.20 s) + (1/2) * (-9.8 m/s^2) * (2.20 s)^2
y ≈ 5.8 m + 19.16 m - 9.71 m
y ≈ 15.25 m

Therefore, the ball clears the wall by approximately 15.25 meters vertically.

(c) Find 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:

x = v0x * t

where x is the horizontal distance, v0x is the initial horizontal velocity, and t is the time taken.

To find v0x, the initial horizontal velocity, we can use the equation:

v0x = v * cos(θ)

Given: v = 10.91 m/s (as found in part (a)), θ = 53.0°, and t = 2.20 s
Plugging in these values, we get:

v0x = 10.91 m/s * cos(53.0°)
v0x ≈ 6.21 m/s

Now we can find the horizontal distance by plugging the values into the equation:

x = (6.21 m/s) * (2.20 s)
x ≈ 13.65 m

Therefore, the horizontal distance from the wall to the point on the roof where the ball lands is approximately 13.65 meters.

C) x= V_x(t) should actually be x= V_x * (t) * (cosine(53))

that should give the the correct answer for x

Then subtract that answer from 24. That is your final answer.

Actually, I meant subtract 24 from x. That should give you the right answer.