Acart 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..

velocity is 13.1 m/s and at an angle of 90 degree with horizontal with respect to cart

Explanation:

As we know that cart speed is uniform and it is moving at 30 m/s

so here we can say that

now the cart is displaced by d = 80 m

so here the time taken by cart too move above distance

now in the same time the projectile must return to the cart so its displacement in y direction must be zero

so here horizontal speed and vertical speed would be

here velocity with respect to cart is 13.1 m/s and it is at 90 degree with horizontal

Speed (relative to the cart) :

angle (to the horizontal) : θ = 23.962°

Projectile motion is a motion by particles that move along a curved path and are affected by gravity  

The velocity in the horizontal direction is constant  

The velocity in the vertical direction changes linearly (the acceleration is constant= g)  

•Horizontal and vertical displacement  

•Total time  

•Time to reach the maximum height (h):  

•The maximum height of projectile  

A cart is moving with constant speed of 30 m/s

so

so the total time taken by cart  

Time of flight = total time , then

Because :

and

we can make tan θ =

θ = 23.962°

velocity position vector

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resultant velocity

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The position of a training helicopter

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Keywords : a cart, projectile, speed, horizontal, vertical

answer; /// i believe that the correct answer is (c)

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