A Particle is projected with a velocity 39.2 m/sec at an angle of 30o to an inclined plane (inclined at an angle of 45 degrees to the horizontal). Find the range on the incline
(a) when it is projected upward
(b) when it is projected downward.
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Time of flight will be same in both cases because the acceleration perpendicular to the plane is same. Therefore,
0 = 39.2 sin 30o t - (½) g cos 45o t2
Or, t = (2×39.2 sin 30)/(g cos 45) = 4√2 sec
(a) Range upward
= 39.2 cos 30o t - (½) g sin 30o t2
= 39.2 × √3/2 × 4√2- (1/2) × 9.8 × (1/2) × (4√2)2 = 113.7m
(b) Range downward
= 39.2 cos 30o × t + (½) g sin 30o t2
= 39.2 × √3/2 × 4√2 + (1/2) × 9.8 × (1/2) × (4√2)2 = 270.5m Ans.