Question

A cart is moving horizontally along a straight line with constant speed 30m/s. A projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved 80m . At what speed (relative to cartt) must the projectile launched? g=10

Answer 1

Case 1 : cart

30 * t = 80

t = 8/3s

Answer 2

Case 2 ; projectile

T = 2 * u * sinx/g

8/3 = 2*u *sinx/10

Answer 3

i'm not really sure...but this is what i got so far:
first, we know Range R = 80m and Time of flight T = 8/3 sec.
let u be the initial speed.
so using their respective formulae, i got
ucosx = R/T
ucosx = 30.................since R/T = 30
u = 30/cosx
the speed of projectile relative to that of cart will be
(speed of cart) - (speed of projectile)...........right?
= 30 - 30/cosx
= 30(1 - secx)
is this right?

Answer 4

The options are 10 10sqrt8 40/3 none of these

Answer 5

@Vaidehi09

Answer 6

lol..yes, i know. we still need to find x.

Answer 7

ok

Answer 8

hw do we find that

Answer 9

that's what i'm trying to work out.

Answer 10

@waterineyes is there a flaw in my approach? if yes, then can u please point it out?

Answer 11

Wait, we don't necessarily have to find x. Assuming from the problem, you're firing the ball straight up while moving on the cart. What we should be focusing on is the vertical velocity, since that is the speed RELATIVE TO THE CART. We want to find how it takes for the ball to reach its maximum height. We use the formula:\[Vy - Gt = 0\], G = 10, to do this. Since we are talking about the maximum height, that means the ball has take half of its time to fall back into the cart. Thus, our time should be 4/3. Doing so yields \[Vy = 40/3\]. Thus


The answer is 40/3.

Answer 12

i dont think that is correct if he throws vertically upwards how the projectile will cover 80m

Answer 13

range will be 0 then...

Answer 14

yes, there should be an angle of projection.

Answer 15

Perhaps the question is unclear to this, but it seems that the cart, because the cart is moving, will move with 30 m/s as well.

Answer 16

It does say the cart is moving.

Answer 17

so the projectile will have intial speed = speed of cart + muzzzle power

Answer 18

if we go by what @dominusscholae says, then the eq should be Vy sinx = gt
so, Vy sinx = 40/3

Answer 19

initial horizontal velocity of the projectile will be equal to 30ms^-1?

Answer 20

Yes. The problem is that it doesn't say whether you're shooting upward. You have to make the assumption that it is, and I made it on the basis of your work. The speed in the x-direction, though, would only come from the cart, while the speed in the y-direction comes from the muzzle shot.

Answer 21

we need to find the angle of projection!

Answer 22

i think the answer is 'none of these'....

Answer 23

I don't know. Is that all the question gives? Far as I know, there's a ton of interpretations and right answers that relate to this question if there is. What is confusing is that in case 1, you didn't factor in a horizontal velocity. Thus I was led to assume we were shooting it up.

Answer 24

case 1 is for the cart. it has velocity only in one direction - horizontal.

Answer 25

and the horizontal velocity of the projectile will be 30ms as well?

Answer 26

i don't think so.

Answer 27

It would have a horizontal velocity of 30 ms. You also have to consider that the problem asked for the speed relative to the cart, which means that the only velocity to find is the vertical velocity of the projectile.