Question

A projectile is thrown with velocity of 50 m/s towards an inclined plane from ground such that it strikes the inclined
plane perpendicularly. The angle of projection of the projectile is 53° with the horizontal and the inclined plane is
inclined at an angle of 45° to the horizontal.
(a) Find the time of flight.
(b) Find the distance between the point of projection and the foot of
inclined plane.

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Answer 1

Correct Answer - (a) t = 7 sec t=7sec, (b) 175 m 175m v x = u x = 50 cos 53 ∘ = 30 m / s vx=ux=50cos53∘=30m/s at striking point v y = − u x = − 30 m / s vy=-ux=-30m/s (a) form time of flight v y = u y + a y t ⇒ − 30 = 40 + ( − 10 ) t ⇒ t = 7 sec vy=uy+ayt⇒-30=40+(-10)t⇒t=7secltbrltgt (b) for distance distance = O Q − M Q = u x t − ( u y t + 1 2 a y t 2 ) =OQ-MQ=uxt-(uyt+12ayt2) distance = 30 × 7 − [ 40 × 7 + 1 2 × ( − 10 ) × 7 2 ] =30×7-[40×7+12×(-10)×72] distance = 210 − ( 280 − 245 ) ⇒ d i s tan c e =210-(280-245)⇒distance= 175 m`