How Do You Calculate the Velocity of a Ball Thrown from a Moving Train?

[leafsfan2]

Homework Statement

A man on a moving train throws a ball at a parked car. Find velocity of the ball when it hits the car if:
train is moving 21 m/s North East;
the man is walking 3.0 m/s [E35S] (relative to train)
ball is thrown 10.0 m/s [W22S] (relative to man)

Homework Equations

V_BG = V_BM + V_MT + V_TG
(B=ball, M=man, T=train, G=ground)

The Attempt at a Solution

I know how to do it if they were all either North or south, but the different directions with the angles confuses me. I've done a bunch of practice questions with only 2 other relative motions (Vxy=Vxz+Vzy), where I use trigonometry to solve for the resultant, but I don't know how to solve it with three relative motions solving for the fourth. I attempted to use trig, but got nowhere with it, and I'm assuming that's not what I should use. My guess is that I should use components? But so far all I wrote was this:
V_BG = V_BM + V_MT + V_TG

 

[SammyS]

You use trig to find the components, don't you?

I take it that E35S means 35° South of due East, etc.

Just find the x-component & y-component for each velocity vector & add them. Then find the magnitude & direction of the resultant. (Usually the positive x-axis points East, & the positive y=axis points North.)

The result looks to me like it will be pretty strange. Also the person throwing the ball must be a contortionist and left handed.

 

[leafsfan2]

Components completely slipped my mind. Thank you, it worked. I got 12m/s [E49N] so that does sound a bit odd but it's just a random question so I'm sure it doesn't matter if it's realistic

 

[SammyS]

Yes, that's a reasonable answer.

The train is moving at more than twice the speed he throws the ball, and he throws the ball in nearly the opposite direction of the train's motion.

 

[rwisz]

Vx = 10.0cos22 - 3.0sin35 + 21cos45
Vy = -10.0sin22 + 3.0cos35 + 21sin45

To solve this problem, we first need to break down the velocities into their x and y components. Using the given information, we can calculate the x and y components of the ball's velocity (VBx and VBy) as well as the man's velocity (VMx and VMy) relative to the ground.

VBx = 10.0cos22 - 3.0sin35
VBy = -10.0sin22 + 3.0cos35
VMx = 3.0cos35 + 21cos45
VMy = 3.0sin35 + 21sin45

Next, we can use the relative motion equation VBG = VBM + VMT + VTG to find the velocity of the ball relative to the ground (VBG).

VBGx = VBx + VMx
VBGx = 10.0cos22 - 3.0sin35 + 3.0cos35 + 21cos45
VBGx = 10.0cos22 + 21.0cos45
VBGx = 26.4 m/s

VBGy = VBy + VMy
VBGy = -10.0sin22 + 3.0cos35 + 3.0sin35 + 21sin45
VBGy = -10.0sin22 + 24.9
VBGy = 14.9 m/s

Finally, we can use the Pythagorean theorem to calculate the magnitude and direction of the ball's velocity when it hits the car.

VBG = √(VBGx² + VBGy²)
VBG = √(26.4² + 14.9²)
VBG = 30.4 m/s

θ = tan⁻¹(VBGy/VBGx)
θ = tan⁻¹(14.9/26.4)
θ = 30.5°

Therefore, the velocity of the ball when it hits the car is 30.4 m/s at an angle of 30.5° below the horizontal.