Launching a projectile while

[idlackage]

Launching a projectile while on something with velocity

Homework Statement

A man is riding on a cart which is moving at a constant velocity of 10 m/s. He wishes to throw a ball through a stationary hoop 5.0 m above his extended hand in such a manner that the ball will be at the top of its trajectory as it passes through the hoop. He throws the ball with a speed of 13 m/s with respect to himself.

a) What must be the vertical component of the initial velocity of the ball?
b) How many seconds after he releases the ball will it pass through the hoop?
c) At what horizontal distance in front of the hoop must he release the ball?

Homework Equations

V₁ = V₂ - a⌂t
⌂d = V₁⌂t + 0.5a⌂t²

The Attempt at a Solution

I pretty much have nothing but that diagram. I have no idea how to even start--I attempted making sense of a triangle calculation somewhere with the givens, but it didn't work out.

Parts b) and c) would be done using one of the 'relevant equations' since I'm guessing I'd obtain the necessary information to plug into them after doing part a).

Thanks for any help.

 

[hotvette]

How would you go about solving the problem if the cart wasn't moving?

 

[idlackage]

How would you go about solving the problem if the cart wasn't moving?

I would start by a component analysis, but since I don't have the angle I can't really do that--which means I'll have to start by finding the angle, but I'm under the impression that all equations I will be using will require components. I know I'm missing some concept but I can't figure it out so it's kind of paradoxical to me right now.

 

[hotvette]

1. The first (and most important) thing you need are the kinematic equations for a projectile. Do you have those? They should be in your book or you should have instruction on how to derive them. You'll have four: x, v_x, y, v_y.

2. Starting with y, you know v_y at the max height and you know the y distance traveled (from the problem statement). These are two equations with two unknowns that can be solved for t and v_y0.

3. You also know v₀ from the problem statement. Using the pythagorean theorem you can find v_x0.

4. Now that you know t and v_x0 you should be able to calculate x.

 

[rwisz]

Based on the given information, we can use the equations of motion to solve for the initial vertical velocity, the time it takes for the ball to pass through the hoop, and the horizontal distance in front of the hoop where the ball should be released.

a) To find the initial vertical velocity, we can use the equation v1 = v2 - at, where v1 is the final vertical velocity (which we want to be 0 m/s at the top of the trajectory), v2 is the initial vertical velocity, a is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes for the ball to reach the top of its trajectory. Since the ball is thrown with a speed of 13 m/s with respect to the man, we can use the Pythagorean theorem to find the initial vertical velocity as follows:

v2^2 = (13 m/s)^2 - (10 m/s)^2

v2 = 6.6 m/s

Therefore, the vertical component of the initial velocity of the ball is 6.6 m/s.

b) To find the time it takes for the ball to pass through the hoop, we can use the equation ⌂d = v1⌂t + 0.5a⌂t^2, where ⌂d is the vertical distance (5.0 m in this case), v1 is the final vertical velocity (which we want to be 0 m/s at the top of the trajectory), a is the acceleration due to gravity (9.8 m/s^2), and ⌂t is the time it takes for the ball to reach the top of its trajectory. Solving for ⌂t, we get:

⌂t = (v1 ± √(v1^2 - 4(0.5)(-9.8)(5.0)))/(-9.8)

⌂t = (0 ± √(6.6^2 + 4(0.5)(9.8)(5.0)))/(-9.8)

⌂t = (0 ± 8.6)/(-9.8)

⌂t = 0.88 s or -0.88 s

Since we are only concerned with the positive value of ⌂t, the time it takes for the ball to pass