A particle is fired with velocity u making an angle $\theta$ with the horizontal.

Question

What is the change in speed at the highest point?

What is important to notice here, is that the change in speed is asked, not velocity. If it was velocity, the answer is $usin\theta$ (which is way too easy)

For speed, directions aren’t important, so initial speed is u and speed at the top is 0. So change in speed should be $u-0=u$. But obviously it’s wrong(other wise I wouldn’t be here). The right answer is $ucos\theta - u$ which seems a bit peculiar to me, so can you please explain what’s going on?

Thanks!

Answer 1

The initial velocity is $\vec{u}=(u\cos\theta,\,u\sin\theta),$ so that the initial speed is $$|\vec{u}|=\sqrt{(u\cos\theta)^2+(u\sin\theta)^2}=|u|.$$ At the highest point, $\vec{u}$ becomes $$(u\cos\theta,\,0),$$ and the corresponding speed here is $$\sqrt{(u\cos\theta)^2+0^2}=|u||\cos\theta|.$$ Thus, the change in speed is $$|u||\cos\theta|-|u|,$$ or since $u> 0$ and $0<\theta<π/2,$

$$u\cos\theta-u.$$

Answer 2

Hint.

It is a projectile.
At the highest point vertical component of velocity becomes $0$, but the horizontal component remains unchanged throughout the flight.