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Formulas/physics/Kinematics/Projection at Angle from Height — Time of Flight

Projection at Angle from Height — Time of Flight

Time of flight when projected at angle θ from height h above ground.
Class 11Class JEE
Derivation

The situation

A body is launched from a height hh above the ground with speed uu at angle θ\theta above the horizontal. It lands on the ground below. How long is it in the air?

This is different from the standard projectile (launched from ground level) — the body starts higher and must fall an extra hh before landing.

Setting up

Take the launch point as the origin. Upward is positive.

  • Initial vertical velocity: uy=usin⁡θu_y = u\sin\theta
  • Vertical acceleration: −g-g
  • At landing, the body is at height −h-h relative to the launch point (it is hh below)

So the vertical displacement at landing is y=−hy = -h.

Derivation

Apply the second equation of motion to the vertical direction:

y=uyt+12(−g)t2y = u_y t + \frac{1}{2}(-g)t^2

At landing, y=−hy = -h and t=Tt = T:

−h=usin⁡θ⋅T−12gT2-h = u\sin\theta \cdot T - \frac{1}{2}gT^2

Rearranging:

12gT2−usin⁡θ⋅T−h=0\frac{1}{2}gT^2 - u\sin\theta \cdot T - h = 0

Multiply through by 22:

gT2−2usin⁡θ⋅T−2h=0gT^2 - 2u\sin\theta \cdot T - 2h = 0

This is a quadratic in TT. Using the quadratic formula with a=ga = g, b=−2usin⁡θb = -2u\sin\theta, c=−2hc = -2h:

T=2usin⁡θ±4u2sin⁡2θ+8gh2gT = \frac{2u\sin\theta \pm \sqrt{4u^2\sin^2\theta + 8gh}}{2g}

T=usin⁡θ+u2sin⁡2θ+2ghg\boxed{T = \frac{u\sin\theta + \sqrt{u^2\sin^2\theta + 2gh}}{g}}

We take the positive root — time must be positive.

Why the formula is more complex than standard projectile

For the standard projectile (launched from ground), the net vertical displacement is zero. That simplification gave T=2usin⁡θgT = \frac{2u\sin\theta}{g}.

Here, the body must additionally fall height hh. The extra 2gh2gh under the square root accounts for this additional fall. As h→0h \to 0, the formula reduces to the standard result:

T→usin⁡θ+u2sin⁡2θg=usin⁡θ+usin⁡θg=2usin⁡θgT \to \frac{u\sin\theta + \sqrt{u^2\sin^2\theta}}{g} = \frac{u\sin\theta + u\sin\theta}{g} = \frac{2u\sin\theta}{g}

Consistent.

Special case: launched horizontally (θ=0\theta = 0)

T=0+0+2ghg=2ghg=2hgT = \frac{0 + \sqrt{0 + 2gh}}{g} = \frac{\sqrt{2gh}}{g} = \sqrt{\frac{2h}{g}}

Recovers the horizontal projection result exactly.

Special case: launched downward at angle θ\theta below horizontal

Replace θ\theta with −θ-\theta (since the initial vertical velocity is now downward):

T=−usin⁡θ+u2sin⁡2θ+2ghgT = \frac{-u\sin\theta + \sqrt{u^2\sin^2\theta + 2gh}}{g}

Note
There is no closed-form "nice" formula for $T$ in the general case — the answer always involves a quadratic. Always set up the vertical displacement equation $y = u_y t - \frac{1}{2}gt^2$ with the correct displacement and solve the quadratic. Do not try to memorise the quadratic formula result — derive it fresh each time.