A cart moves with velocity vC = 10 m/s in the horizontal direction as shown below.

https://online-s.physics.uiuc.edu/cg...04/01/cart.gif

A ball is ejected from the cart with an initial velocity vB with respect to the cart such that an observer on the ground sees the ball move directly vertical. The ball reaches its maximum height at a time t = 2 s after it is launched. The next two questions pertain to this situation.

The angle θ (as measured by someone on the cart) is greater than 90 degree.

how can I get the angle measured?

What is vB, the magnitude of the initial velocity of the ball with respect to the cart?

vB = 10 m/s
vB = 18 m/s
vB = 22 m/s
vB = 26 m/s
vB = 30 m/s

In order for the ball to go straight up relative to the ground it must have been given a backwards horizintal component relative to the cart of vC. So V_h = horizintal velocity relative to cart = -vC. Since its maximum altitude is reached in 2 seconds, its initial upward velocity can be found from:
v = u + at
where u is the initial upward velocity. So at its maximum height you have:
0 = u -gt
Solve for u. Then the initial velocity of the ball relative to the car can be found by applying Pythagorian Theorem to the initial horizontal and verical velocity components to get the magnitude of vB. The value of θ is determined from Arctan(V_h/u).

Hi, I wonder what is V_h/u

I just wonder that why the horizontal velocity relative to cart is -vC not +vC?

Quote:

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Originally Posted by western50

Hi, I wonder what is V_h/u

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V_h = horizontal velocity = -vC. It is a negative number because the ball must be thrown backwards out of the cart.

u = initial vertical velocity.

What will be the difference between the angle 0, measured by someone on the cart and measured by someone on the ground?

Try to visualise the situation. If you were on the cart, you'll see the projectile going behind you, right?

But if you were on the ground, you said that an observer sees the projectile as going directly upwards. Hence, theta will here by 90 degrees.

Is that okay?