A circus performer stands on a platform and throws an apple from a height of 49 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 = 30 m/s,
how far from the end of the platform should the second performer stand? (Assume θ = 30°.) the answer key is 51.5m. If you need to see the figure , give me your email and I will send it to you as fast as I could
1st:-1/2*9.81t^2+(30*sin(-30)t)+49-->-4.9t^2-15t+49
2nd: use the quadratic formula to find time in second-->(-(-15)+or -square root (-15-4(-4.9*49))divide the whole thing by -9.81=-5.04s or 1.98s
3rd: take 1.98s multiply for the answer--> 30cos(-30)*1.98=51.5m
Okay Anonymous, I don't know who you are
but you don't need to see the figure then show me how to solve the problem. Your resonse doesn't help in anyway on solving this question.
To find the distance from the end of the platform where the second performer should stand, we can use projectile motion equations.
First, let's break down the initial velocity v0 of the apple into its horizontal and vertical components.
The horizontal component (v0x) is given by:
v0x = v0 * cos(θ)
The vertical component (v0y) is given by:
v0y = v0 * sin(θ)
In this case, θ is given as 30°, so we can substitute the given values into the equations above:
v0x = 30 m/s * cos(30°)
v0y = 30 m/s * sin(30°)
Now, let's calculate the time it takes for the apple to reach the ground. We'll consider the vertical motion of the apple since the horizontal motion is constant.
Using the equation:
y = y0 + v0yt - (1/2)gt^2
Where:
y = -49 m (since it is measured downward from the initial position of the apple)
y0 = 0 m (initial vertical position)
v0y = 30 m/s * sin(30°)
g = 9.8 m/s^2 (acceleration due to gravity)
t = ? (time)
-49 = 0 + (30 * sin(30°))t - (1/2)(9.8)t^2
Now, we can solve this quadratic equation to find the time (t). We simplify and rearrange the equation:
-49 = (15)t - (4.9)t^2
Rearranging the equation, we get:
4.9t^2 - 15t - 49 = 0
Now, we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. In this case, using the quadratic formula is the most straightforward approach.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Using our equation:
a = 4.9, b = -15, c = -49
we can substitute these values into the quadratic formula and solve for t. Note that we will take the positive value of t since it represents the time it takes for the apple to reach the ground.
t = [15 ± √((-15)^2 - 4(4.9)(-49))] / (2 * 4.9)
After evaluating this expression, we find t ≈ 3.15 seconds.
Now that we have the time it takes for the apple to hit the ground, we can calculate the horizontal distance it has traveled.
Using the equation:
x = v0xt
Where:
x = ? (horizontal distance)
v0x = 30 m/s * cos(30°)
t = 3.15 s (time)
We substitute the given values into the equation:
x = (30 * cos(30°)) * 3.15
Evaluating this expression, we find x ≈ 51.5 meters.
Therefore, the second performer should stand approximately 51.5 meters from the end of the platform to catch the apple.