Question

a particle is thrown at time t=0, with a velocity of 10m/s at an angle of 60 with the horizontal, from a point on an inclined plane making an angle 30 with the horizontal. The time when the velocity of the projectile becomes parallel to the incline is

Answer 1

The velocity of projection u = 10m/sthe angle of projection θ =60°At time t the angle made by the projectile with the horizontal is 30°which is parallel to the inclined.Horizontal velocity after time t is given by, u co60° = V cos30°or V = u co60°cos30°=10×22×3√=103√. the angle at which inclined plane is inlined with the horizontal ϕ=30°let at time t the projectile becomes parallel to inclined plane.so , 103√ sin30°=10 sin60°−10×tor 103√×12=10×3√2−10 tor 10 t= 5(3√−13√)=103√or t = 13√ s
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Answer 2

Answer:1/✓3

Explanation:

See,

First we find the total time

T= 2usin(A-B)/gcosB

​where A is the angle the projectile makes with the horizontal and B is the angle that the inclined plane makes with the horizontal.

Therefore using A as 60 and B as 30

T= 2/root 3

Now it says PARALLEL to the inclined plane,

if we make a basic diagram if a projectile motion along inclined plane , we see that at the MIDDLE, the velocity is TAGENTIAL to the curve and therefore parallel to the incline.

Hence req time=t/2

which is 1/root 3 sec.