Man on a Flatcar With Ball

[postfan]

Homework Statement

A person is riding on a flatcar traveling at a constant speed v1= 20 m/s with respect to the ground. He wishes to throw a ball through a stationary hoop in such a manner that the ball will move horizontally as it passes through the hoop. The hoop is at a height h=4 m above his hand. He throws the ball with a speed v2= 18 m/s with respect to the flatcar. Let g=10 m/s2 and neglect air drag completely. (see figure)

(a) At what horizontal distance x in front of the hoop must the person release the ball? (in meters)

(b) When the ball leaves his hand, what is the direction of the velocity vector of the ball as seen from the flatcar? (angle αcar with respect to the horizontal in degrees)

(c) When the ball leaves his hand, what is the direction of the velocity vector of the ball as seen from the ground? (angle αground with respect to the horizontal in degrees

Homework Equations

The Attempt at a Solution

I found the y component of the initial velocity using the equation 0=v_0*2=2(-10)(4), which I found to be 4sqrt(5). I used this to find the x component of the initial velocity which I calculated to be 2sqrt(61). I found the angle with respect to the boxcar by taking sin^-1=4sqrt(5)/18), angle =29.795 degrees.

To find the time to reach the hoop I used the equation 4=4sqrt(5)-5t^2, t=.8944. Knowing the time and x-component of the velocity I calculated the horizontal distance to be 13.971.

I have no clue how to find the angle relative to the boxcar.

Can someone please check my work on the first 2 parts and how to get the third part? Thanks!

 

[kosovo dave]

Could you post your equations in symbolic form? I'm not sure where 0=v_0*2=2(-10)(4) is coming from. Are you saying that twice the initial velocity is equal to zero?

 

[postfan]

I used the equation v=v_0^2+2*a*x for (1) and x=v_0t+.5t^2 for (2).

 

[Simon Bridge]

Homework Statement

A person is riding on a flatcar traveling at a constant speed v1= 20 m/s with respect to the ground. He wishes to throw a ball through a stationary hoop in such a manner that the ball will move horizontally as it passes through the hoop. The hoop is at a height h=4 m above his hand. He throws the ball with a speed v2= 18 m/s with respect to the flatcar. Let g=10 m/s2 and neglect air drag completely.

It sounds like you have been using set equations without considering the physics of the problem.
Otherwise you'd know how to find the angles.

The following exercise should help you understand what to do:
if the person throws at speed u=v2 (I like to minimize subscripts) and angle θ wrt the flatcar, and the flatcar is moving at speed v=v1 wrt the ground
then :
1. what is the initial horizontal speed $u_x$ wrt the flatcar?
2. what is the initial vertical speed $u_y$ wrt the flatcar?
3. what is the initial horizontal speed $u_{xg}$ wrt the ground?
4. what is the initial vertical speed $u_{yg}$ wrt the ground?
5. what is the initial speed $u_g$ wrt the ground?
6. what is the initial angle $\theta_g$ wrt the ground?
... leave everything as variables - you are writing out equations.

It is best practice to to the algebra before you do the substitutions. If nothing else it makes it easier to troubleshoot your equations. In order to help you we need to see your reasoning too.

 

[Nickie]

Your work for the first two parts seems to be correct. For the third part, you can use the fact that the angle of the velocity vector with respect to the ground is equal to the sum of the angle with respect to the flatcar and the angle of the flatcar with respect to the ground. In this case, the angle of the flatcar with respect to the ground is 0 degrees since it is moving horizontally. Therefore, the angle of the velocity vector with respect to the ground is also 29.795 degrees.