A projectile is launched from ground level with a speed of 25 m/s at an angle of 42° to the horizontal. What is the horizontal distance from the starting point of the projectile when it hits the ground?

Resolve the vertical velocity of the projectile (25sin42) and use v=u+at where v=-25sin42m/s, u=25sin42m/s, a=-9.8m/s/s to find t. t= (50sin42)/9.8= 3.41sResolve horizontal velocity (25cos42) and multiply by time (3.41) to find distance travelled.3.41*25cos42= 63.4m