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?
a.35.8 m/s at 24 degrees
b.38.6 m/s at 54 degrees
c.27 m/s at 35 degrees
d.24 m/s at 44 degrees

Answer 1

i mean what part u did not get

Answer 2

do u have any idea how to start

Answer 3

With all due respect, is it possible you may have copied a 35.8 m/s instead of 32.8 m/s for option A? I feel this is the correct response if this is the case for the following reasons. First, we know that the horizontal velocity of the cart is a constant 30 m/s and during a given time interval covers 80 m before the ball returns to this cart. Knowing this fact, we can use the kinematic equation\[\Delta x = v _{i} \Delta t + \frac{ 1 }{ 2 } a \Delta t ^{2}\]
In the horizontal direction, there is no acceleration by either the ball or cart as there is no external force acting horizontally on either to cause an acceleration to take place (this is all on the assumption there is no friction or air resistance). As a result, a portion of our equation resolves itself to zero. Thus, we are left with\[\Delta x = v _{i} \Delta t\]
If we re arrange this equation and solve for 'Delta t' we find the total time of flight to be 2.7 s from the time the projectile is fired until the projectile lands back in the cart. Knowing that 1.35 s into its flight, the projectile will reach its apex in its trajectory, the vertical velocity will be 0 m/s at this point. With this fact in mind, we can use the equation \[v _{f} = v _{i} + a \Delta t\]
For the vertical (y) direction and solve for the initial velocity of the object in the vertical direction alone. Consequently we would find\[v _{i} = v _{f} - a \Delta t = 0 - (-9.81)(1.35) = 13.2 m/s\]
Again, this velocity is in the upward direction. Now, we can determine the resultant velocity by using the fact that the horizontal velocity is 30 m/s and the initial velocity in the vertical direction is 13.2 m/s. According to pythagoreans theorem\[R = \sqrt{13.2^{2}+30^{2}} = 32.8 m/s\]
To determine the angle of launch, we now consider the tangent function.\[\Theta = \tan ^{-1}(\frac{ Opposite }{ Adjacent }) = \tan ^{-1}(\frac{ 13.2 }{ 30 }) = 24^{o}\]
Your final answer then would read 32.8 m/s @ 24 degrees above the horizontal axis. This is why I ask that you consider if a typo has been made. I hope this helps.