A circus performer throws an apple toward a hoop held by a performer on a platform (see figure below). The thrower aims for the hoop and throws with a speed of 28 m/s. At the exact moment the thrower releases the apple, the other performer drops the hoop. The hoop falls straight down. (Assume
d = 27 m and h = 54 m.
Neglect the height at which the apple is thrown.)
(a) At what height above the ground does the apple go through the hoop?
m
Steve can you explain what is the square root of 5 at the 28/square root of 5 is?
To find the height above the ground at which the apple goes through the hoop, we can use the equations of motion.
Let's break down the problem:
Given:
- Initial speed of the apple (v) = 28 m/s
- Distance from the thrower to the hoop (d) = 27 m
- Height of the hoop (h) = 54 m
We need to find the height (y) at which the apple goes through the hoop.
Using the equations of motion, we can calculate the time it takes for the apple to reach the hoop.
Step 1: Calculate the time of flight (t) using the vertical motion equation:
h = (1/2)gt^2
54 = (1/2)(9.8)t^2
108 = 9.8t^2
t^2 = 108/9.8
t^2 = 11.02
t ≈ √11.02
t ≈ 3.32 s
Step 2: Calculate the horizontal distance covered (x) by the apple using the horizontal motion equation:
x = v * t
x = 28 * 3.32
x ≈ 92.96 m
Step 3: Calculate the height at which the apple goes through the hoop (y) using the vertical motion equation:
y = d - x
y = 27 - 92.96
y ≈ -65.96 m
The height at which the apple goes through the hoop is approximately -65.96 m above the ground.
Note: The negative sign indicates that the apple is below the starting point at this height.
To find the height at which the apple goes through the hoop, we can analyze the motion of the apple and the hoop separately.
Let's first look at the motion of the apple. Since we're neglecting the height at which the apple is thrown, we can assume its initial vertical position is at the same height as the hoop, which is 54 m.
Using the information given, we can find the time it takes for the apple to reach the hoop by using the equation of motion:
d = vit + (1/2)at²
where:
d = distance traveled (27 m, which is the horizontal distance between the thrower and the hoop)
vi = initial velocity (in the vertical direction, this is 0 m/s)
t = time taken to reach the hoop
a = acceleration (due to gravity, this is -9.8 m/s²)
Since the initial velocity in the vertical direction is 0 m/s, the equation simplifies to:
d = (1/2)at²
Substituting the given values:
27 m = (1/2)(-9.8 m/s²)t²
Solving for t²:
t² = (2 * 27 m) / -9.8 m/s² = -5.51 s²
Since time cannot be negative, we disregard the negative sign:
t² = 5.51 s²
Taking the square root of both sides:
t ≈ 2.35 s
The time it takes for the apple to reach the hoop is approximately 2.35 seconds.
Now, we can find the vertical distance the apple travels in this time by using the equation of motion:
vf = vi + at
where:
vf = final velocity (0 m/s, as the apple reaches its peak and starts falling down)
vi = initial velocity (which is 0 m/s because the apple was dropped originally)
a = acceleration (due to gravity, this is -9.8 m/s²)
t = time taken (2.35 s)
Simplifying the equation, we have:
0 m/s = 0 m/s + (-9.8 m/s²) * 2.35 s
Solving for the vertical displacement:
d = vit + (1/2)at²
d = 0 m/s * 2.35 s + (1/2) * (-9.8 m/s²) * (2.35 s)²
d ≈ -27.54 m
The negative sign indicates that the apple falls below the initial height.
Therefore, the height above the ground at which the apple goes through the hoop is approximately 54 m - 27.54 m = 26.46 m above the ground.
I assume that d is the horizontal distance from the thrower to the spot below the hoop.
so, throwing it directly toward the hoop's initial position, the angle θ is such that
tanθ = 54/27 = 2
So, the horizontal speed is
vx = 28 cosθ = 28/√5 = 12.52 m/s
So, it takes the apple 27/12.52 = 2.156 seconds to reach the falling hoop.
Now, the hoop at time 2.156 has fallen to a height of
54 - 4.9*2.156^2 = 31.223 m
extra credit: do the constraints work out? What is the apple's height at t=2.156?