Question

Show that the speed with which a body is projected upwards is equal to the speed with which it returns to the point of projection.

Answer 1

Let $\sf{v}$ be the speed of a projectile when it reaches a particular point in its trajectory after a time $\sf{t.}$

Let $\sf{u}$ be the initial speed of the projectile thrown with angle of projection $\theta.$

So the horizontal and vertical components of the initial velocity are $\sf{u\cos\theta}$ and $\sf{u\sin\theta}$ respectively.

• $\sf{u_x=u\cos\theta}$

• $\sf{u_y=u\sin\theta}$

By first equation of motion,

$\longrightarrow\sf{v_x=u_x+a_xt}$

Since there's no horizontal forces acting on the projectile,

$\longrightarrow\sf{v_x=u\cos\theta}$

And,

$\longrightarrow\sf{v_y=u_y+a_yt}$

Since the projectile experiences acceleration due to gravity downwards,

$\longrightarrow\sf{v_y=u\sin\theta-gt}$

We consider upward motion as positive.

Thus the speed $\sf{v}$ is given by,

$\longrightarrow\sf{v=\sqrt{(v_x)^2+(v_y)^2}}$

$\longrightarrow\sf{v=\sqrt{(u\cos\theta)^2+(u\sin\theta-gt)^2}}$

$\longrightarrow\sf{v=\sqrt{u^2\cos^2\theta+u^2\sin^2\theta+g^2t^2-2ugt\sin\theta}}$

Since $\sf{\sin^2\theta+\cos^2\theta=1,}$

$\longrightarrow\sf{v=\sqrt{u^2+g^2t^2-2ugt\sin\theta}\quad\quad\dots(1)}$

If $\sf{v}$ is the speed of the projectile attained when it returns to the plane of projection, then $\sf{t}$ will be time of flight, given by,

• $\sf{t=\dfrac{2u\sin\theta}{g}}$

Hence (1) becomes,

$\longrightarrow\sf{v=\sqrt{u^2+g^2\left(\dfrac{2u\sin\theta}{g}\right)^2-2ug\sin\theta\cdot\dfrac{2u\sin\theta}{g}}}$

$\longrightarrow\sf{v=\sqrt{u^2+4u^2\sin^2\theta-4u^2\sin^2\theta}}$

$\longrightarrow\sf{v=\sqrt{u^2}}$

$\longrightarrow\underline{\underline{\sf{v=u}}}$

∴ The projectile returns to its plane of projection with the same speed as that of its initial speed.