Question

A particle is projected horizontally with a speed u from the top of a plane inclined at an angle theta with the horizontal how far from the point of projection will the partical strike the plane ?​

Answer 1

Take X,Y-axes as shown in attached figure.

Suppose that the particle strikes the plane at a point P with coordinate (x,y). Consider the motion between A and P.

$\sf{ \red{Motion~ in~ x-direction:}} \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{Initial~ velocity = u}} \\ \: \ \: \: \: \sf{ \blue{Acceleration= 0}} \\ \: \: \: \: \: \: \: \: \: \: \: \ \: \sf{ \blue{x = ut}} \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)$

$\sf{ \red{Motion~ in~ y-direction:}} \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{Initial~ velocity = 0}} \\ \: \ \: \: \: \sf{ \blue{Acceleration= g}} \\ \: \: \: \: \: \: \: \: \: \: \: \ \: \sf{ \blue{y = \frac{1}{2} {gt}^{2} }} \: \: \: \: \: ....(ii)$

$\maltese \: \sf{Eliminating \: \: t \: \: from \: (i) \: and \: (ii)}$

$\longrightarrow\: \: \: \: \: \: \: \: \: \sf {y = \frac{1}{2} g \frac{ {x}^{2} }{ {u}^{2} } } \\ \\ \: \: \: \: \: \: \: \: \: \: \longrightarrow\sf{ \pink{y = \tan\theta}}$

Thereafter,

$\implies \sf { \frac{ {gx}^{2} }{ {2u}^{2} }} = x \tan\theta \: \\ \\ \implies\sf \bold{x = \frac{2 {u}^{2} \tan \theta }{g}}$

$\sf{Clearly \: the \: point \: P \: corresponds \: to,}$ $\\ ~~~~~~~~~~ \sf{\: x = \: \frac{2 {u}^{2} \: \tan \theta }{g} }$

then,

$\sf{y = x \tan \theta = \frac{2 {u}^{2} { \tan}^{2} \theta }{g} }$

$\sf{The \:distance \: \: AP = } \: l = \sqrt{ {x}^{2} + {y}^{2} } \\ \\ \: \: \: \: \: \: \: \implies \sf{\frac{2 {u}^{2} }{g} tan \theta \sqrt{1 + { \tan}^{2} \theta } } \\ \\ \: : \implies \sf{ \pink{ \frac{2 {u}^{2} }{g} \tan \theta \sec \theta }}$

Answer 2

Explanation:

When a particle is projected horizontally with a speed

�

u from the top of a plane inclined at an angle

�

θ with the horizontal, the distance traveled by the particle along the plane before striking it can be calculated using the equations of motion.

Let's assume that the particle takes time

�

t to strike the plane. During this time, the particle only experiences motion along the horizontal direction. The vertical motion is independent of the horizontal motion because the initial velocity in the vertical direction is zero.

In the horizontal direction, the initial velocity

�

u remains constant, and the distance traveled

�

d can be calculated using the formula:

�

=

�

⋅

�

d=u⋅t

To find the time

�

t, we need to consider the vertical motion of the particle. The vertical motion can be described using the equation:

ℎ

=

1

2

�

�

2

h=

2

1

gt

2

where

ℎ

h is the vertical distance covered and

�

g is the acceleration due to gravity. Since the initial vertical velocity is zero, the above equation simplifies to:

ℎ

=

1

2

�

�

2

h=

2

1

gt

2

The vertical distance

ℎ

h can be expressed as:

ℎ

=

�

⋅

sin

⁡

(

�

)

h=x⋅sin(θ)

where

�

x is the distance along the inclined plane.