Speed and Direction of a Projectile at Any Instant
Physics · Motion In A Plane · NEET
At any moment, a projectile has two velocity parts: horizontal vx = u cos(theta), which never changes, and vertical vy = u sin(theta) - gt, which slows, stops, then reverses. The real speed is v = sqrt(vx^2 + vy^2), and its direction is angle = tan-inverse(vy / vx) from the horizontal. Memory hook: "x-speed is frozen, y-speed fights gravity" — only vy changes, so at the top vy = 0 and speed = vx = u cos(theta).
At every instant the velocity (dashed) is the vector sum of a constant horizontal part vx = u cos(theta) (red) and a changing vertical part vy = u sin(theta) - gt (purple). At the top vy = 0, so the speed equals u cos(theta). Overall speed = sqrt(vx^2 + vy^2), direction = tan-inverse(vy/vx).
Your doubts, answered
Why does the horizontal velocity vx never change during the flight?
After launch, the only force on the projectile is gravity, which pulls straight down. There is no horizontal force (we ignore air resistance). So the horizontal acceleration is zero, and vx = u cos(theta) stays the same from launch to landing. Only the vertical part vy feels gravity and keeps changing.
What is the speed at the highest point? Is it zero?
No. At the highest point only the vertical velocity becomes zero (vy = 0), because gravity has just stopped the upward motion. The horizontal velocity is still there: vx = u cos(theta). So the speed at the top is u cos(theta), not zero. Speed is zero only if you throw the ball straight up (theta = 90 degrees).
How do I find the speed at any time t?
Use two steps. First find the parts: vx = u cos(theta) (constant) and vy = u sin(theta) - g t. Then combine them: speed v = sqrt(vx^2 + vy^2). This works for any instant — going up, at the top, or coming down.
How do I find the direction of the velocity at any instant?
The velocity always points along the tangent to the path. Its angle above (or below) the horizontal is angle = tan-inverse(vy / vx). When vy is positive the ball is rising; when vy is negative it is falling, so the angle points below the horizontal.
Why is the speed the same at the same height on the way up and the way down?
vx is unchanged, and at equal heights the magnitude of vy is equal (gravity gives back exactly what it took away). Since v = sqrt(vx^2 + vy^2) depends only on the sizes of vx and vy, the speed matches. Only the direction flips (upward on the way up, downward on the way down).
⚠️ The NEET trap ✗ Speed at the highest point of a projectile is zero. ✓ At the highest point only vy = 0. The horizontal part vx = u cos(theta) survives, so the speed there is u cos(theta). 🧠 Only the vertical speed dies at the top. The horizontal speed keeps going — the ball is still moving forward.
Real NEET questions
NEET 2023
A ball is projected from point A with a velocity 20 m/s at an angle 60 degrees to the horizontal direction. At the highest point B of the path, the speed v (in m/s) of the ball will be:
A · Zero
B · 10 ✓
C · 20
D · 10*sqrt(3)
Solution: At the highest point the vertical velocity vy = 0, so only the horizontal velocity remains. Speed at top = u cos(theta) = 20 x cos(60) = 20 x 0.5 = 10 m/s. Answer: 10 m/s (option B). Trap: do not pick zero — horizontal speed never dies.
NEET 2022
A ball is projected with a velocity 10 m/s at an angle of 60 degrees with the vertical direction. Its speed at the highest point of its trajectory will be:
A · Zero
B · 5*sqrt(3) m/s
C · 5 m/s ✓
D · 10 m/s
Solution: Angle is given from the vertical. 60 degrees from the vertical = 30 degrees from the horizontal. At the top only the horizontal component stays. Horizontal component = 10 x cos(60 from vertical) = 10 x sin(30 from horizontal) = 10 x 0.5 = 5 m/s. Answer: 5 m/s (option C). Read the angle reference carefully — from vertical, not horizontal.
NEET 2023
The position of a particle is given by r(t) = 4t i + 2t^2 j + 5 k, where t is in seconds and r in metres. Find the magnitude and direction (with respect to the x-axis) of the velocity v(t) at t = 1 s.
A · 3*sqrt(2) m/s, 30 deg
B · 3*sqrt(2) m/s, 45 deg
C · 4*sqrt(2) m/s, 45 deg ✓
D · 4*sqrt(2) m/s, 60 deg
Solution: Velocity is the time derivative of position: v = dr/dt = 4 i + 4t j. At t = 1 s, v = 4 i + 4 j. Magnitude = sqrt(4^2 + 4^2) = sqrt(32) = 4*sqrt(2) m/s. Direction = tan-inverse(vy/vx) = tan-inverse(4/4) = 45 degrees. Answer: 4*sqrt(2) m/s at 45 degrees (option C).
Solved Motion In A Plane NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for the speed of a projectile at any instant?
v = sqrt(vx^2 + vy^2), where vx = u cos(theta) (constant) and vy = u sin(theta) - g t. Substitute the time t and compute the square root.
What is the direction of velocity of a projectile at time t?
The velocity points along the tangent to the path at that point. Its angle with the horizontal is angle = tan-inverse(vy / vx). A positive vy means it is going up; a negative vy means it is coming down.
Is the velocity of a projectile ever horizontal?
Yes, exactly once — at the highest point, where vy = 0 and the whole velocity is horizontal with speed u cos(theta).
Does the speed of a projectile stay constant?
No. Because vy keeps changing due to gravity, the speed changes too. It is largest at launch and landing, and smallest (equal to u cos(theta)) at the highest point.
When is the projectile's speed minimum?
At the highest point of the path. There vy = 0, so v = u cos(theta), the smallest speed during the whole flight.
