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

nop

Answer 2

u were offline at that time!

Answer 3

i posted this once but i didnt understand

Answer 4

The time taken by the cart to travel 80m = 8/3 sec.
This is equal to the time of flight of the projectile = 2u sinx/g = 8/3, where u is the initial velocity and x is the angle at which the projectile is projected.
u*sin x= 40/3 . sin x = 40/3u , cos x = sqrt( 1-sin^2 x ) =
(sqrt( 9u^2-1600))/3u .

The horizontal range of the projectile is given by u^2*sin 2x/g=80m .

u^2*sin 2x = 800 , sin 2x = 2*sin x*cos x = 40(sqrt( 9u^2 - 1600 ))/9u^2

Plugging this in the range equation , we get 40(sqrt( 9u^2 - 1600 ))/9=800,

sqrt(9u^2-1600) = 180 , squaring on both sides and simplifying ,
9u^2=30800 , u= 58.5 m/sec.

Answer 5

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

Answer 6

I think it is none of these.Not sure.Rechecking the values now.Please wait.

Answer 7

ok

Answer 8

i think u r correct

Answer 9

Oops I made a mistake, here is the new solution:

The time taken by the cart to travel 80m = 8/3 sec.

This is equal to the time of flight of the projectile = 2u sinx/g = 8/3, where u is the initial velocity and x is the angle at which the projectile is projected.

u*sin x= 40/3 . sin x = 40/3u , cos x = sqrt( 1-sin^2 x ) = (sqrt( 9u^2-1600))/3u .

The horizontal range of the projectile is given by u^2*sin 2x/g=80m . u^2*sin 2x = 800 , sin 2x = 2*sin x*cos x = 40(sqrt( 9u^2 - 1600 ))/9u^2

Plugging this in the range equation , we get 80(sqrt( 9u^2 - 1600 ))/9=800, sqrt(9u^2-1600) = 90 , squaring on both sides and simplifying , 9u^2= , u= 32,829 m/sec.

Still getting none of the above option.I've checked my calculation.

Answer 10

wat .......32,829 m/sec.

Answer 11

man..........its huge

Answer 12

i think we have to consider relative motion

Answer 13

as it is thrown from the moving cart

Answer 14

Sorry it is supposed to be a point.It is 32.829 m/sec

Answer 15

And yes it is relative to the velocity of the cart.

Answer 16

ok............thxxx