Question

A stone is projected with speed V0 at angle of projection 53° with the horizontal. At the instant the velocity of the stone makes angle 37° with the horizontal, the radius of trajectory of its path is 45(V↓0)²/ng the value of n is​

Answer 1

Answer:

If a stone is thrown with a speed v

0

at an angle θ

0

with horizontal, find the velocity of the stone when its line of motion makes an angle θ with horizontal.

Q2

The angle of projection with the horizontal at which projectile is projected is

Q3

A projectile is fired with speed u at an angle Θ with the horizontal. Its velocity when projectile makes angle β with horizontal

Answer 2

Given info : A stone is projected with speed V₀ at angle of projection 53° with the horizontal. At the instant the velocity of the stone makes angle 37° with the horizontal.

To find : The radius of trajectory of its path is 45V₀²/ng , the value of n is ....

solution : initial velocity, u = v₀cos53° i + v₀sin53° j

u = 3/5v₀ i + 4/5v₀ j .....(1)

at an instant, the velocity of the stone makes angle 37° with the horizontal,

tan37° = vy/vx = 3/4

⇒vy = 3/4 vx

but vx remains same. so vx = 3/5v₀ [ from eq (1).]

now vy = 3/4 × 3/5v₀ = 9/20v₀

so, v = 4/5v₀i + 3/5v₀j

magnitude, |v| = √{(3/5)² + (9/20)²}v₀ = 3/4v₀ .....(2)

now at equilibrium,

centripetal force = weight towards centre of curvature

⇒mv²/R = mgcos37°

⇒(3/4v₀)²/g(4/5)= R

⇒R = (9v₀²)/(16 × 4g/5) = 45v₀²/64g

therefore the value of n = 64.