Question

ONLY FOR GENIUS
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Q.) On an inclined plane of inclination 30 degree A ball is thrown at an angle of 60 with the horizontal from the foot of the incline with velocity of 10√3 m/s . If g=10m/s^-2 then the time in which ball with hit the inclined plane is ?

(A) 1.15 sec

(B) 6 sec

(C) 2 sec

(D) 0.92 sec
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#NEED FULL EXPLANATION★
#10 POINTS★
#NO SPAMS PLEASE★

Answer 1

Formula for time of flight on an inclined plane is :-

T = 2u Sin α /g Cos(theta)

Here ,

α= 60-30=30° ,which is angle of projection from inclined plane.

theta = 30°, which is angle of inclination (of the inclined plane)

∴ T= 2 x 10√3 x 1/2 / 10 x √3/2 (Sin 30°= 1/2 30°=√3/2)

T= √3 / √3/2

= √3 x 2/√3

= 2 seconds

Answer 2

Answer:

Given:

Inclination of inclined plane(alpha) =30°

Angle of projection(beta)=60°-30°=30°

Velocity of projection(u)=10 root3 m/s

Acceleration due to gravity(g)=10m/s^2

To calculate:

Time taken by the ball to hit the inclined plane

Method:

Displacement along y axis(y)=0m at the time of flight(T)

y=UyT+1/2Ay t^2

0=u sin betaT-gcos alpha T^2

T=2u sin beta/g cos alpha

T=2*10 root3*sin30°/10*cos30°

T=10 root3/5 root3

T=2 seconds

Therefore, option (C) is correct.

Hope it will help you

Mark the brainliest