Question

a particle is projected from horizontal surface (at an angle 45° with the horizontal) from a point which is at a normal distance of 2m from the door of the wall. It just falls from the top of the wall and falls on ground 4m from it. Maximum height reached by the particle is​

Answer 1

Explanation:

Range of projectile R=8+4=12 m

Angle of projection θ=45

o

Using equation of trajectory, y=xtanθ(1−

R

x

)

where x=8 m

Or y=8×tan45(1−

12

8

)

⟹ y=

3

8

m

Answer 2

Concept:

Horizontal motion is defined as a projectile motion in an exceedingly horizontal plane depending upon the force engaged on it.

Given

The angle from which the particle is projected with the surface is $45^{\circ}$ and also the particle from the wall to the door is $2m$ and therefore the particle falls from the highest of the wall and falls on the bottom is $4m.$

Find

We have to search out the utmost height reached by the particle.

Given

Firstly, we'll find the range of the particle.

$\begin{aligned}R&=2+4\\ &=6m\end$

Now, we are going to use the formula $y=x\tan \theta \left(1-\frac{x}{R}\right)$.

From the given, $\theta=45,x=2$ and $R=6$, we get

$\begin{aligned}y&=2\times \tan 45^{\circ}\left(1-\frac{2}{6}\right)\\ &=2\times 1\left(1-\frac{1}{3}\right)\\ &=2\left(\frac{2}{3}\right)\\ &=\frac{4}{3}\end$

Hence, the most height reached by $\frac{4}{3}m.$

#SPJ3