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?

Let V = the speed of the projectile relative to the cart., and that is fired at angle x relative to the horizontal.

Because the cart moves horizontally at 30 m/s,
The horizontal launch velocity of the projectile is
u = V*cos(x) + 30

The vertical launch velocity of the projectile is
v =  V*sin(x)

d =  80 m, the distance traveled by the cart when the projectile lands in it.

The time that the cart travels is
t = (80 m)/(30 m/s) = 2.667 s

The projectile reaches maximum height in t/2 = 1.333 s.
Because v - gt =0, therefore
t = v/g
Vsin(x)/9.8 = 1.333   
Vsin(x) = 13.0663                 (1)

Also,
u*t = d, therefore
[Vcos(x) + 30]*2.667 = 80
Vcos(x) + 30 = 80/2.667 =  30
Vcos(x) = 0                         (2)

Because V ≠ 0, cos(x) = 0  => x = 90°
This means that the projectile is fired straight up.

From (1), obtain
V*sin(90) = 13.0663
V = 13.07 m/s (nearest hundredth)

The projectile is fired at 90° relative to the horizontal (straight up), at a velocity of 13.07 m/s.

Ithink the answer is c but then again i'm probably wrong