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.

Asked by sourabhprakash | 10th Sep, 2010, 03:16: PM

Expert Answer:

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.

Answered by  | 10th Sep, 2010, 03:29: PM

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