Question

A cart is moving horizontally with constant velocity vx. The cart then shoots a ball with an initial velocity vo straight up relative to the cart.

a) show that the ball will hit the car regardless of the values of vx or vo.

b)IN terms of vx and vo, what is the horizontal distance traveled by the cart while the ball is in the air?

Answer 1

what have you done so far?

Answer 2

i used the formula x=x(initial) +Vx+ .5at^2, and solved for x and got x=vxt. i substituted d for x and got d=vxt. since the velocity in x for the ball is the same as the cart, d=v(ball)t.

I think this answers the first part of the question but im not sure

Answer 3

your choice of notation is confusing me
if x is the distance the object moves in the x-direction and Vx is the horizontal component of velocity, your formula is wrong
the x component of velocity is *not* affected by gravity, because the acceleration due to gravity is entirely in the vertical direction in these kinds of problems

Answer 4

the .5at^2 can be removed from the equation youre right. so can the initial position. that is why i came to the conclusion of d=vx*t

Answer 5

right and that equation can understand, but i would probably try to invoke Vo into the equations since that is mentioned in the problem, in which case you want to start with something like\[\vec v_o=\vec v_x+\vec v_y=\vec v_x+\vec v_{oy}-\frac12\vec at^2\]

Answer 6

i mean your explanation is okay... but you should specify that the x-component of the ball's velocity in the expression for Vo is the same as the x-component in the velocity of the cart's Vo

Answer 7

okay...

Answer 8

for instance\[\vec v_{0ball}=\vec v_{0xball}+\vec v_{yball}=\vec v_{0xcart}+\vec v_{yball}\]\[\vec d_{ball}=\vec v_{0xcart}t+\vec v_{yball}t\]

Answer 9

then all you need to say is \[\vec d_{ballx}=\vec v_{0cartx}t\]

Answer 10

\[\vec d_{ballx}=\vec v_{0ballx}t=\vec v_{0cartx}t=\vec d_{cart}\]is probably better
anyway my point was just to use the expression for Vo of the ball and cart