Arugby player passes the ball 7.25 m across the field, where it is caught at the same height as it left his hand. at what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used

Let x =  angle at which the ball was thrown, relative to the horizontal.

The vertical component of velocity is
Vy = (12 m/s)sin(x) = 12sin(x) m/s

At maximum height, the vertical velocity is zero. The time to attain maximum height is given by
12sin(x) - gt = 0
t = [12sin(x)]/9.8 = 1.2245sin(x) s

The horizontal component of velocity is 12cos(x) m/s.
The time taken to travel 7.25 m is 2t. Because distance = velocity*time, therefore
12cos(x)*(2*1.2245sin(x)) = 7.25
14.694*(2sin(x)cos(x)) = 7.25

Note that 2sin(x)cos(x) = sin(2x). Therefore
sin(2x) = 7.25/14.694 = 0.4934
2x = arcsin(0.4934) = 29.56°
x = 14.8° (nearest tenth)

 14.8° (approx. 15°)

It may possibly be c i think |-/