there are two friends A & B 15m apart.From A wall is at 5m. A throws a ball to B. At what angle should he project the ball so that it reaches B. Height of wall is 10m.

The trajectory of projectile motion,

y = tanθ x - (gx²/(2V²cos²θ)

Since the ball must cross the wall of 10 m at a distance of 5m from A, (5,10) is the point on the trajectory.

10 = 5tanθ  - (125/V²cos²θ)     ... (1)

Also the range is, 15m.

R = 2V²sinθcosθ/g

75 = V²sinθcosθ

1/V²= sinθcosθ/75

Using this value of 1/V² in (1),

10 = 5tanθ  - (125tanθ/75)    

10 = 5tanθ  - (5tanθ /3)   

tanθ = 3

θ = tan^-1 3 = 71°

Regards,

Team,

TopperLearning.