there are two A & B friends 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, 03:15: PM
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
Using this value of 1/V2 in (1),
10 = 5tanθ - (125tanθ/75)
10 = 5tanθ - (5tanθ /3)
tanθ = 3
θ = tan-1 3 = 71°
Answered by | 10th Sep, 2010, 03:28: PM
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