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Question: A stone is projected horizontally from a point P, so that it hits the inclined plane perpendicularly...

A stone is projected horizontally from a point P, so that it hits the inclined plane perpendicularly. The inclination of the plane with the horizontal is θ\theta and the point P is at a height h above the foot of the incline, as shown in the figure. Determine the velocity of projection.

A

v0=2gh1+cot2θv_0 = \sqrt{\frac{2gh}{1+cot^2\theta}}

B

v0=2gh2+cot2θv_0 = \sqrt{\frac{2gh}{2+cot^2\theta}}

C

v0=2gh3+cot2θv_0 = \sqrt{\frac{2gh}{3+cot^2\theta}}

D

v0=2gh4+cot2θv_0 = \sqrt{\frac{2gh}{4+cot^2\theta}}

Answer

(D) v0=4gh2+cot2θv_0 = \sqrt{\frac{4gh}{2 + cot^2 \theta}}

Explanation

Solution

Here's how to determine the velocity of projection:

  1. Set up coordinate system: Let the point of projection P be the origin (0,0). Let the x-axis be horizontal to the right and the y-axis be vertically downwards.

  2. Equations of motion:

    • Initial velocity: u⃗=v0i^\vec{u} = v_0 \hat{i} (assuming projection is to the right, towards the incline).
    • Acceleration: a⃗=gj^\vec{a} = g \hat{j}.
    • Position at time tt: x(t)=v0tx(t) = v_0 t, y(t)=12gt2y(t) = \frac{1}{2} g t^2.
    • Velocity at time tt: vx(t)=v0v_x(t) = v_0, vy(t)=gtv_y(t) = g t.

  3. Perpendicular impact condition:

    • The stone hits the inclined plane perpendicularly. The slope of the velocity vector is mv=vyvx=gtv0m_v = \frac{v_y}{v_x} = \frac{g t}{v_0}.
    • The inclined plane makes an angle θ\theta with the horizontal and goes downwards to the right. So its slope is mplane=−tan⁡θm_{plane} = -\tan \theta.
    • For perpendicular impact, mv=−1mplanem_v = -\frac{1}{m_{plane}}.
    • gtv0=−1(−tan⁡θ)=cot⁡θ\frac{g t}{v_0} = -\frac{1}{(-\tan \theta)} = \cot \theta.
    • This gives gt=v0cot⁡θ…(1)g t = v_0 \cot \theta \quad \ldots(1).

  4. Geometry of the setup:

    • The point P is at a height hh above the foot of the incline (F). Let the horizontal distance from P to the vertical line passing through F be LL.
    • In the right triangle formed by P, F, and the point vertically below P on the horizontal line of F, we have tan⁡θ=hL\tan \theta = \frac{h}{L}.
    • So, L=hcot⁡θL = h \cot \theta.
    • The equation of the inclined plane, passing through F (L,h)(L, h) and having a slope of −tan⁡θ-\tan \theta, is Y−h=(−tan⁡θ)(X−L)Y - h = (-\tan \theta)(X - L).

  5. Point of impact:

    • The stone hits the plane at (x,y)(x, y). Substituting x=v0tx = v_0 t and y=12gt2y = \frac{1}{2} g t^2 into the plane equation:
    • 12gt2−h=(−tan⁡θ)(v0t−L)\frac{1}{2} g t^2 - h = (-\tan \theta)(v_0 t - L).
    • Substitute L=hcot⁡θL = h \cot \theta:
    • 12gt2−h=(−tan⁡θ)(v0t−hcot⁡θ)\frac{1}{2} g t^2 - h = (-\tan \theta)(v_0 t - h \cot \theta)
    • 12gt2−h=−v0ttan⁡θ+htan⁡θcot⁡θ\frac{1}{2} g t^2 - h = -v_0 t \tan \theta + h \tan \theta \cot \theta
    • 12gt2−h=−v0ttan⁡θ+h\frac{1}{2} g t^2 - h = -v_0 t \tan \theta + h
    • 12gt2=2h−v0ttan⁡θ…(2)\frac{1}{2} g t^2 = 2h - v_0 t \tan \theta \quad \ldots(2).

  6. Solve for v0v_0:

    • Substitute t=v0cot⁡θgt = \frac{v_0 \cot \theta}{g} from (1) into (2):
    • 12g(v0cot⁡θg)2=2h−v0(v0cot⁡θg)tan⁡θ\frac{1}{2} g \left(\frac{v_0 \cot \theta}{g}\right)^2 = 2h - v_0 \left(\frac{v_0 \cot \theta}{g}\right) \tan \theta
    • 12gv02cot⁡2θg2=2h−v02cot⁡θtan⁡θg\frac{1}{2} g \frac{v_0^2 \cot^2 \theta}{g^2} = 2h - \frac{v_0^2 \cot \theta \tan \theta}{g}
    • v02cot⁡2θ2g=2h−v02g\frac{v_0^2 \cot^2 \theta}{2g} = 2h - \frac{v_0^2}{g}
    • Multiply by 2g2g:
    • v02cot⁡2θ=4gh−2v02v_0^2 \cot^2 \theta = 4gh - 2v_0^2
    • v02cot⁡2θ+2v02=4ghv_0^2 \cot^2 \theta + 2v_0^2 = 4gh
    • v02(2+cot⁡2θ)=4ghv_0^2 (2 + \cot^2 \theta) = 4gh
    • v0=4gh2+cot⁡2θv_0 = \sqrt{\frac{4gh}{2 + \cot^2 \theta}}.