Question

A projectile is fired with a speed u at an angle θ above a horizontal field. The coefficient of restitution of collision between the projectile and the field is e. How far from the starting point, does the projectile makes its second collision with the field?

Answer 1

Thanks for asking the question!

ANSWER::

Projected velocity = u

Angle of projection = Ф

Velocity will be same as u , when projectile hits the ground.

Component of velocity is parallel to ground , u cos Ф should remain constant.

Vertical component of projectile undergoes a change after collision.

e = u sin Ф / v

v = eu sin Ф

Now,

For 2nd projectile motion

Velocity of projection = U = √[(u cosФ)² + (eu sinФ)²]

Angle of projection = α = tan⁻¹ (eu sinФ / a cosФ) = tan⁻¹(e tanФ)

tan α = e tanФ

y = x tan α - (gx² sec²α) / 2u² [Equation 1]

Here , y = 0

tan α = e tan Ф

sec²α = 1 + e² tan²Ф

u² = u²cos²Ф + e²sin²Ф

Putting values in Equation 1

x e tanФ = gx²( 1 + e² tan²Ф) / 2u²(cos²Ф + e² sin²Ф)

x = 2eu² tanФ(cos²Ф + e² sin²Ф) / g( 1 + e² tan²Ф)

x = (2eu² tanФ - cos²Ф ) / g

x = eu²sin2Ф / g

From starting point O , it will fall at a distance of = u²sin2Ф/g + eu²sin2Ф/g

= u²sin2Ф(1 + e) / g

Hope it helps!