A circus performer stands on a platform and throws an apple from a height of 50 m above the ground with an initial velocity
v0
as shown in the figure below. A second, blindfolded performer must catch the apple. If
v0 = 21 m/s,
how far from the end of the platform should the second performer stand? (Assume θ = 30°.) The answer is 41.8. Can somebody shows me step by step to the answer
1st:use this format equation--> -1/2*9.81t^2+(21*sin(-30)t)+50 which gives -->-4.9t^2-10.5t+50
2nd: now use the quadratic formula to find the time-->(-(-10.5)+/- the square root of(-10.5^2-(4*-4.9*50)))divide by(2*-4.9)=-4.44 and2.30
3rd: use only the positive value 2.30 and multiply with 21cos(-30)--> 21cos(-30)*(2.30)=41.8m
The answer key is 41.8 but I don't know how to solve for this question so that I can get the same answer.
The height of the ball is given by
h(t) = 50 + 21/2 t - 4.9t^2
Assuming the catcher is on the ground, you need
h(t) = 0, so t=4.44
So. Now you know that horizontal speed of the apple is
21 cos30° = 18.186 m/s
So, it will travel 4.44 * 18.186 = 80.75 meters.
Hmmm. Either the length of the platform is 38.95 meters long, or the catcher is not on the ground.
Have I misinterpreted the situation, having no diagram?
Is there anyway for I to send you the picture of the figure so that you can have a better understanding and help me with this question
To find the distance from the end of the platform where the second performer should stand, we can use the equations of motion for projectile motion.
Step 1: Split the initial velocity into its horizontal and vertical components.
Since the angle θ is given as 30°, we can find the horizontal and vertical components of the initial velocity using trigonometry.
The horizontal component, v0x, can be found using the equation:
v0x = v0 * cosθ
The vertical component, v0y, can be found using the equation:
v0y = v0 * sinθ
Step 2: Find the time taken for the apple to reach the ground.
To find the time taken for the apple to reach the ground, we can use the vertical motion equation:
y = v0y * t - (1/2) * g * t^2
In this case, y = -50 m (negative because it is measured downward) and g = 9.8 m/s^2 (acceleration due to gravity).
Set the equation to zero (since the apple reaches the ground), and solve for t.
Step 3: Find the horizontal distance traveled by the apple.
To find the horizontal distance traveled by the apple, we can use the horizontal motion equation:
x = v0x * t
Substitute the value of t obtained in Step 2 and calculate x.
Step 4: Calculate the distance from the end of the platform.
Since the second performer needs to stand at the point where the apple will fall, the distance from the end of the platform is equal to the horizontal distance traveled by the apple.
Plug in the values of v0, θ, and g into the equations, and calculate the distance from the end of the platform.
Using the given values:
v0 = 21 m/s
θ = 30°
g = 9.8 m/s^2
Let's do the calculations:
Step 1:
v0x = v0 * cosθ = 21 m/s * cos(30°) = 18.18 m/s
v0y = v0 * sinθ = 21 m/s * sin(30°) = 10.50 m/s
Step 2:
y = v0y * t - (1/2) * g * t^2
-50 m = 10.50 m/s * t - (1/2) * 9.8 m/s^2 * t^2
Rearrange the equation to:
0 = -4.9 * t^2 + 10.50 * t - 50
Use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
a = -4.9, b = 10.50, c = -50
Calculating t gives two solutions:
t1 = 4.964 s
t2 = 2.134 s
Since the time taken cannot be negative, we use t = 2.134 s as the time taken for the apple to reach the ground.
Step 3:
x = v0x * t = 18.18 m/s * 2.134 s = 38.80 m
Step 4:
The distance from the end of the platform where the second performer should stand is equal to the horizontal distance traveled by the apple, which is 38.80 m. Rounded to one decimal place, the answer is approximately 41.8 m.