Time, range and height

0 = 25\sin 40^\circ\,t - 4.9t^2

T = \frac{25\sin 40^\circ}{4.9} = 3.28\text{ s}

R = 25\cos 40^\circ \times 3.279\ldots = 62.8\text{ m}

H = \frac{(25\sin 40^\circ)^2}{19.6} = 13.2\text{ m}

the greatest height comes halfway, where the vertical velocity is zero

Projectiles

A kicked ball does two things at once.

It moves forwards and it rises and falls, and the two don't interfere. Split its velocity in two and each part has a simple story of its own.

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Two arrows

The ball leaves at 25 m s⁻¹ at 40°. Its velocity splits into

\begin{aligned} 25\cos 40^\circ &= 19.15\ldots \text{ across} \\ 25\sin 40^\circ &= 16.06\ldots \text{ up} \end{aligned}

At the top the upward arrow is gone, but the ball still moves across at $19.15\ldots$ m s⁻¹.

Only gravity

Nothing pushes the ball sideways, so the horizontal arrow never changes. Gravity pulls it down, so the vertical arrow shrinks by $9.8$ every second: to zero at the top, then growing downwards until it lands, a mirror image of the climb.

Same landing, higher

Now kick it at 50° instead. It climbs far higher and stays up longer, yet lands in exactly the same place. The range is

R = \frac{U^2\sin 2\alpha}{g}

and $\sin 80^\circ = \sin 100^\circ$ .

The longest kick

$\sin 2\alpha$ is greatest, $1$ , when $\alpha = 45^\circ$ . Halfway between 40° and 50°, the dashed kick goes furthest of all:

\frac{U^2}{g} = \frac{625}{9.8} = 63.8\text{ m}

Chapter 3

Across at a steady pace. Up and down under gravity.

Throws from cliffs, speed and direction mid-flight, the general formulae and where they come from, the equation of the path, and twelve graded exam questions with full worked solutions are waiting.

Read the full chapter › Jump to the questions ›