Question

A body thrown along a frictionless inclined plane of angle of inclination 30 degrees covers a distance 40m along the plane. If the body is projected with the same speed at an angle of 30 degree with the ground, it will have a range of(take g=10m/s)

Answer 1

First by using the first part of the question, we have to find the initial velocity which the body was thrown (u).

Here we can use Energy conservation law to solve this,

If we take the Potential energy is 0 when the body was thrown from the bottom of the inclined plane we can write,

Kinetic Energy when the body was thrown from the bottom of the inclined plane = Potential Energy when the body stops.

1/2 x m x u^2 = m x g x 40Sin(30)

u = 20 m/s

Then we use this initial velocity to find the answer by using Linear Motion Equations,

We can first apply s = ut + 1/2at^2 for vertically upward direction,

0 = 20Sin(30) x t + 1/2 x (-10) x t^2

0 = 10t - 5t^2

By solving this quadratic equation we can get,

5t(2-t) = 0

t(2-t) = 0

t = 0 / t = 2

This gives the two answers for t where s = 0,

• Before throwing the body s = 0 , t = 0
• After the body landed on the surface s = 0 , t = 2

So we should take t = 2

Then we can apply s = ut + 1/2at^2 horizontally ---> direction,

s = 20Cos(30) x 2 + 1/2 x 0 x 2^2 , there is no acceleration for that direction

s = 20√3/2

s   = 20√3

Answer 2

Answer:

The range will be 20√3m

Explanation:

Given:

Inclination(θ)=30°

Distance(s)=40m

The speed is constant(v)=0m/s

To find: Range=?

Solution:
The effective acceleration can be calculated using,

$a=gsin$θ

=10*sin30

=10*0.5=5ms⁻²

We know that, the third equation of motion is

v²=u²+2as

0=u²+2*5*40

∴u²=400

∴u=20

As we know,

R$=\frac{u^{2}sin2 }{g}$θ = u²sin60/g=(400*√3/2)/10=20√3.

Therefore, the range will be 20√3m