Is velocity zero at the top of a projectile's path?
Short answer: Only the vertical part becomes zero; the horizontal part u cosθ never changes.
The trap
The wrong idea
“At the top of a projectile's path its velocity is zero.”
It sounds reasonable, and a lot of JEE/NEET aspirants carry it into the exam. Here's a one-question check, then the explanation that makes it click.
Quick check
Try it before you peek
At the highest point of an angled projectile (no air), its velocity is:
- Zero
- Horizontal, u cosθ
- Vertical, u sinθ
Show the answer
B. Horizontal, u cosθ
Only the vertical part becomes zero; the horizontal part u cosθ never changes.
The click
Split projectile motion into two independent parts. Launch with speed u at angle θ above the horizontal: the horizontal component is u cos θ and the vertical component is u sin θ. With no air, the only force is gravity, and it acts vertically.
Horizontally there is no force, so no acceleration: the horizontal velocity stays u cos θ for the entire flight. Vertically, gravity reduces the vertical velocity by g every second until it reaches zero; that instant is the highest point, reached at t = u sin θ / g.
So at the top only the vertical part is zero. The velocity there is u cos θ, pointing horizontally, and it is the minimum speed of the whole flight. The acceleration at the top is still g downward, perpendicular to the velocity at that moment.
The velocity is zero at the top only for a throw straight up (θ = 90°), where cos θ = 0. Any angled projectile is still moving sideways at its peak, which is why it carries on forward and lands somewhere else. A related fact worth remembering: the kinetic energy at the top is ½m(u cos θ)², not zero.
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