Horizontally Projected Body (Projectile from a Height)
Physics · Motion In A Plane · NEET
When a body is thrown horizontally from a height, its horizontal velocity stays constant (u) while it falls freely under gravity in the vertical direction. Time of flight depends ONLY on the height: t = sqrt(2H/g), and horizontal range R = u x sqrt(2H/g). Memory hook: "Horizontal is lazy, vertical is falling" - side motion never changes, downward motion speeds up just like a dropped stone.
A body launched horizontally with speed u keeps vx = u throughout while vy = gt grows downward. The path curves into a parabola; it lands at range R = u x sqrt(2H/g), striking the ground at angle theta where tan theta = sqrt(2gH)/u.
Your doubts, answered
Does the horizontal velocity change while the body falls?
No. There is no force acting in the horizontal direction (air resistance is ignored), so the horizontal velocity stays the same value u for the whole flight. Only the vertical velocity grows because gravity pulls down. So at every instant vx = u and vy = g t.
A ball is thrown horizontally and another is dropped from the same height at the same moment. Which lands first?
They land at the SAME time. The horizontal motion has no effect on the falling. Both have zero initial vertical velocity and fall the same height H, so both take t = sqrt(2H/g). The thrown ball just lands farther away, not later. This is the famous independence of horizontal and vertical motion.
How do I find the time of flight for a horizontally projected body?
Use only the vertical motion. Initial vertical velocity is 0, so H = (1/2) g t^2. Rearranged: t = sqrt(2H/g). The horizontal speed u never appears. Example: from H = 20 m with g = 10, t = sqrt(2x20/10) = sqrt(4) = 2 s.
Why does the horizontal speed not affect the time to fall?
Time in the air is decided by how long it takes to fall the vertical height H. Vertical and horizontal motions are independent, so changing u only changes how FAR it goes sideways (the range), never how long it stays up. This is a top NTA trap.
What is the velocity of the body when it hits the ground?
Combine the two components. Horizontal vx = u (unchanged). Vertical vy = g t = sqrt(2gH). Resultant speed v = sqrt(u^2 + 2gH). The angle below the horizontal is given by tan(theta) = vy / vx = sqrt(2gH) / u.
⚠️ The NEET trap ✗ Thinking the ball thrown harder (larger u) stays in the air longer or hits the ground later. ✓ Time of flight t = sqrt(2H/g) depends ONLY on height H and g. A larger u increases the range, not the time. A thrown ball and a dropped ball from the same height land together. 🧠 More speed = more distance sideways, SAME time down.
Real NEET questions
2019
Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings 100 m apart and of the same height 200 m, with the same speed 25 m/s. When and where will the two bullets collide? (g = 10 m/s^2)
A · after 2 s at a height of 180 m ✓
B · after 2 s at a height of 20 m
C · after 4 s at a height of 120 m
D · they will not collide
Solution: Step 1 (horizontal): Both bullets move towards each other, closing the 100 m gap at 25 + 25 = 50 m/s. Time to meet t = 100 / 50 = 2 s. Step 2 (vertical): Both start with zero vertical velocity and fall identically, so they stay at the same height as each other and DO collide. Drop in 2 s: h = (1/2) g t^2 = (1/2)(10)(2^2) = 20 m. Step 3: Height above ground = 200 - 20 = 180 m. Answer: after 2 s at a height of 180 m (option A). Key idea: horizontal closing decides WHEN, free fall decides the HEIGHT.
Solved Motion In A Plane NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Is the path of a horizontally projected body a straight line?
No, it is a parabola. The equation of the path is y = (g / (2 u^2)) x^2, which is the equation of a parabola. Constant horizontal speed plus accelerating vertical fall bends the path into a curve.
What is the formula for horizontal range from a height?
R = u x t = u x sqrt(2H/g), where u is the horizontal launch speed and H is the height. Unlike ground-to-ground projectiles, here there is no angle term because the launch is purely horizontal.
At what angle does the body strike the ground?
tan(theta) = vy / vx = sqrt(2gH) / u, measured below the horizontal. As the height H increases, vy grows and the strike angle gets steeper.
Does mass affect a horizontally projected body?
No. Both the fall and the horizontal motion are independent of mass (air resistance ignored). A heavy and a light body thrown with the same u from the same height follow the same path and land together.
