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Question: A particle projected from ground moves at angle \({{45}^{\circ }}\) with horizontal one second after...

A particle projected from ground moves at angle 45∘{{45}^{\circ }} with horizontal one second after projection and speed is minimum two seconds after the projection. The angle of projection of the particle is (neglect the air resistance).

Explanation

Solution

To solve this question one needs to know some things about projectile motion. Using the knowledge of the motion of a projectile, form some equations in terms of the angle of projectile and the initial velocity then solve the equation to calculate the angle of projection.

Formula used:
vy=usin⁡θ−gt{{v}_{y}}=u\sin \theta -gt
where vy{{v}_{y}} is velocity of the projectile in vertical direction at time t, g is acceleration due to gravity, u is initial velocity and θ\theta is angle of projection.
vx=ucos⁡θ{{v}_{x}}=u\cos \theta
where vx{{v}_{x}} is velocity of the projectile in horizontal direction at time t.
T=2usin⁡θgT=\dfrac{2u\sin \theta }{g}
where T is the time of flight of the projectile.

Complete step by step answer:
It is given that the speed of the projectile is minimum after two seconds and we know that the speed of the projectile is minimum at the highest point from the ground. A projectile projected from the ground, takes half of the time equal to its time of flight to reach the maximum height.
This means that 2=T2=(2usin⁡θg)22=\dfrac{T}{2}=\dfrac{\left( \dfrac{2u\sin \theta }{g} \right)}{2}.
⇒usin⁡θg=2\Rightarrow \dfrac{u\sin \theta }{g}=2 …. (i).
It is given that at time t=1st=1s, the motion of the projectile makes an angle of 45∘{{45}^{\circ }} with the horizontal.

Let the speed of the projectile at this time be v. Therefore, its speed in the horizontal direction is vx=vcos⁡45∘=v2{{v}_{x}}=v\cos {{45}^{\circ }}=\dfrac{v}{\sqrt{2}} and its speed in the vertical direction is vy=vsin⁡45∘=v2{{v}_{y}}=v\sin {{45}^{\circ }}=\dfrac{v}{\sqrt{2}}
But we know that after one second, vx=ucos⁡θ{{v}_{x}}=u\cos \theta and vy=usin⁡θ−g(1){{v}_{y}}=u\sin \theta -g(1). Therefore, from the above four equations we get that vx=vy=ucos⁡θ=usin⁡θ−g{{v}_{x}}={{v}_{y}}=u\cos \theta =u\sin \theta -g.
This means that g=usin⁡θ−ucos⁡θg=u\sin \theta -u\cos \theta
⇒u=gsin⁡θ−cos⁡θ\Rightarrow u=\dfrac{g}{\sin \theta -\cos \theta }
Substitute the value of u in equation (i).
⇒(gsin⁡θ−cos⁡θ)sin⁡θg=2\Rightarrow \left( \dfrac{g}{\sin \theta -\cos \theta } \right)\dfrac{\sin \theta }{g}=2
⇒sin⁡θ=2(sin⁡θ−cos⁡θ)\Rightarrow \sin \theta =2(\sin \theta -\cos \theta )
⇒sin⁡θ=2sin⁡θ−2cos⁡θ)\Rightarrow \sin \theta =2\sin \theta -2\cos \theta )
On simplifying further we get sin⁡θcos⁡θ=2\dfrac{\sin \theta }{\cos \theta }=2
⇒tan⁡θ=2\Rightarrow \tan \theta =2
∴θ=tan⁡−12\therefore \theta ={{\tan }^{-1}}2

Therefore, the angle of projection is equal to tan⁡−1(2){{\tan }^{-1}}(2).

Note: The motion of the projectile is influenced by the gravitational force only (in absence of air resistance). Since the direction of the gravitational force is always in the downward direction, the projectile is only accelerated in the downward direction and has a constant velocity along the horizontal direction.If you do not remember the formula for the time of flight, then you can easily derive it using kinematic equations.