Question

A cart is moving horizontally along a straight line with constant speed of 30 m/s. A projectile is fired from the moving cart in such a way that it will return to the cart after the cart has moved 80 m. At what speed (relative to the cart) and at what angle (to the horizontal) must the projectile be fired?

I asked this question earlier but was not getting it right..

Answer 1

i guess the value of v0 is 30m/s, right?

Answer 2

The projectile, say a ball, is thrown up into the air and it falls back into the cart when it has travelled 80m. Since it is travelling at 30 m/s, it should take 8/3 s to fall back, meaning that the ball is air-borne for 8/3 s. If you recall that the time of flight of a vertically thrown body t=2u/g, it is easy to see that the launch velocity u=gt/2 =40/3 m/s. That deals with vertical velocity.
Now about horizontal velocity, horizontal velocity vx doesn’t change with time since there is no gravity horizontally and no wind either. So, it should be same as cart velocity.
Now, you know both the horizontal and vertical velocities.
To calculate v= (vx^2 + vy^2)^1/2.
Angle of projection = arctan (vy/vx)

Answer 3

how did you arrive at the 8/3s

Answer 4

the cart is travelling with 30 m/s and it has travelled for 80 m. So it took a time of 80/30 s to travel which is 8/3 s. the projectile was thrown at the beginning of the journey and it falls back at the end of the journey. So it is in the air for 8/3 s too.

Answer 5

And about vx, it doesn't change with time if your reference frame is the ground. If it was the cart, it would always be above the cart, and hence vx from the cart would appear to be zero.

Answer 6

@shivam_bhalla what is the value of v0?

Answer 7

v0 is the intial velocity with which the projectile is thrown at

Answer 8

By v0, if you mean the velcity of projectile, you can calculate it by evaluating the root of the sum of squares of vx and vy.
v0=(vx^2 + vy^2) ^1/2

Answer 9

@Prof.Grace , Is Theta = tan^-1 (4/9) in your answer

Answer 10

well, i'm still stuck and have not reached there

Answer 11

Do you have the final answer of the question or not?

Answer 12

no, not yet

Answer 13

I meant that is the final answer for the question given to you or not??

Answer 14

V_0 = g*t/2

Answer 15

V_0 = g*t/2 --> Why and How??

Answer 16

vy is gt/2. Not v0.

Answer 17

There is nothing about air resistance, so I neglected it. Horizontal initial velocity is equal to that of the cart, therefore theta is Pi/2. Time is 8/3 second. V_y = 0 at t/2. Therefore, V_0*sin(Pi/2)= g*t/2.

Answer 18

@Prof.Grace i think angle of projection with horizontal is 60.14degree and relative velocity of projection is 60.26m/s

Answer 19

OK logically nsutton sounds right. For the person in the cart, it appears to him that the ball is thrown straight up into the air and it comes back to him

So
v_x = 0
v_y = v_0 -gt

t = 80/30 = 8/3
We get
Time to reach maximum height = 4/3

Therefore
we know that

v_f = u_0 -g ( t)
At mmaximum ht v_f = 0

0 = v_0-gt/2
v_0=gt/2
v_0= 10*(4/3) = 40/3

and theta = 90 :)

This is wrt cart.

For wrt ground


Angle wr.t ground =
tan (theta)= (40/3)/ 30 = 4/9
theta = tan^(-1) (4/9)

And Initial velocity wrt ground
v= 30 i + (40/3) j

@experimentX
@mos1635
@Siddharth18
Please check this solution

Answer 20

Question asked for with respect to cart I thought. But I agree with the with respect to ground also.

Answer 21

@Aditya790 is right
@nsutton 's final comment is right

Answer 22

the angle must be zero. the horizontal component of velocity must be equal to that of cart ... and since the projectile if fired from ground, it will automatically have this horizontal component.

Also if we consider the the cart as our frame of reference, it is same as 1d projectile case.

Answer 23

the angle must be 90 (to the orizontal)

Answer 24

relavent to the cart...

Answer 25

@mos1635 is right.

Answer 26

looks I should have said, zero to vertical ... LOL

Answer 27

since the cart is moving at constant velocity with respect to the ground,the acceleration of the cart wrt ground =0.therefore we can consider the cart as an inertial frame of reference.now i am doing the calculations of velocity and acc.of the projectile with respect to the cart.
now since the projectile is falling back on the cart,therefore the cart and the projectile cover the same horizontal distance in same time and as such their horizontal velocities will be the same.now t=8/3 is also the time of flight of the projectile so