Q.The ceiling of a long hall is $25\ \text{m}$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40\ \text{m s}^{-1}$ can go without hitting the ceiling of the hall?

Concept: Projectile motion. The ceiling caps the maximum height, which caps the launch angle; the largest allowed angle gives the largest range (since it's below $45^\circ$).

1. Maximum height formula: $H = \dfrac{u^2\sin^2\theta}{2g} = 25\ \text{m}$, with $u=40\ \text{m/s}$, $g=9.8\ \text{m/s}^2$.
2. Solve: $\sin^2\theta = \dfrac{2gH}{u^2} = \dfrac{490}{1600} = 0.30625 \implies \theta \approx 33.6^\circ$, $\cos\theta \approx 0.8329$. …

Setting up

The launch angle is free to choose, but the peak of the ball's trajectory must never exceed the $25\ \text{m}$ ceiling. A larger launch angle gives a higher peak, so the ceiling sets an upper limit on the angle that can be used.

Step 1 — Find the limiting angle from the height cap

Maximum height: $H = \dfrac{u^2\sin^2\theta}{2g}$. With $H = 25\ \text{m}$, $u = 40\ \text{m/s}$, $g = 9.8\ \text{m/s}^2$:

$\sin\theta \approx 0.5534 \implies \theta \approx 33.6^\circ, \qquad \cos\theta = \sqrt{1-0.30625} \approx 0.8329$ …

Concept: Work Directly with the Velocity Components, Never Solve for the Launch Angle

Method: Constrained-Components Route ($v_x^2+v_y^2=u^2$ and $v_y^2=2gH$), Range from $R=\dfrac{2v_xv_y}{g}$ — No $\theta$, No Inverse Trig at All

The existing solutions first find the limiting launch angle $\theta\approx33.6^\circ$ from the height cap, then substitute that angle into the range formula. This method skips the angle entirely: it treats the vertical and horizontal velocity components themselves as the unknowns, pins one of them down from the ceiling constraint, gets the other from $u$, and plugs both directly into a component-only form of the range formula — no $\sin\theta$, $\cos\theta$, or $\theta$ ever appears.

Step 1 — The two components are linked by the launch speed

For any launch angle, the horizontal and vertical components of the initial velocity satisfy:

This is just Pythagoras on the velocity triangle — true regardless of what angle was actually used.

Step 2 — The ceiling fixes the vertical component directly

The maximum height reached depends only on the vertical component (the horizontal motion never affects how high the ball rises):

So $v_y = \sqrt{490} \approx 22.14\ \text{m/s}$ — using all of this component (grazing the ceiling exactly) gives the ball the most horizontal reach possible, since any smaller $v_y$ wastes launch speed that could have gone into $v_x$ instead, while any larger $v_y$ hits the ceiling.

Step 3 — The horizontal component is whatever's left

From Step 1:

Step 4 — Range from the components directly

The range formula $R=\dfrac{u^2\sin2\theta}{g}$ can be rewritten using $\sin2\theta=2\sin\theta\cos\theta$ and $v_x=u\cos\theta$, $v_y=u\sin\theta$: …