KinematicsHard

Jumping the ditch (projectile motion)

A motorcyclist rides up a ramp inclined at α = 35° to jump a ditch d = 8 m wide. Find: minimum take-off speed, maximum height of the jump, and the angle of the velocity with the horizontal after 0.25 s.

Given data

α = 35°d = 8 mt = 0.25 sg = 9.81 m/s²

Review the theory: Cinematica

Steps

0 / 3

The minimum take-off speed makes the rider land exactly on the far edge of the ditch: the range must equal d. For a projectile launched and landing at the same height the range is d = v₀²·sin(2α)/g; impose d and solve for v₀.
Write your working — the numerical result must appear in the text m/s
At the highest point the vertical velocity vanishes (it rises until gravity stops it). Only the initial vertical component v₀sinα sets the height reached; by energy conservation in the vertical: h = v₀²·sin²α/(2g).
During the flight the horizontal component vₓ stays constant (no horizontal force), while the vertical one decreases due to gravity: v_y = v₀sinα − g·t. The angle of the velocity with the horizontal is the ratio of the two components: θ = arctan(v_y/vₓ).

Full worked solution

The minimum take-off speed makes the rider land exactly on the far edge of the ditch: the range must equal d. For a projectile launched and landing at the same height the range is d = v₀²·sin(2α)/g; impose d and solve for v₀.
$v_0 = \sqrt{\frac{d\,g}{\sin 2\alpha}} = \sqrt{\frac{8\cdot 9.81}{\sin 70°}}$
$v_0 = \sqrt{78.5/0.940} \approx \mathbf{9.14\,m/s}$ .
At the highest point the vertical velocity vanishes (it rises until gravity stops it). Only the initial vertical component v₀sinα sets the height reached; by energy conservation in the vertical: h = v₀²·sin²α/(2g).
$h = \frac{v_0^2\sin^2\alpha}{2g}$
$h = \frac{83.5\cdot0.329}{19.62} \approx \mathbf{1.40\,m}$ .
During the flight the horizontal component vₓ stays constant (no horizontal force), while the vertical one decreases due to gravity: v_y = v₀sinα − g·t. The angle of the velocity with the horizontal is the ratio of the two components: θ = arctan(v_y/vₓ).
$\theta = \arctan\!\frac{v_0\sin\alpha - g t}{v_0\cos\alpha}$
$\theta = \arctan(2.79/7.49) \approx \mathbf{20.4^\circ}$ .

Result:

v_0 \approx 9.14

m/s,

h \approx 1.40

\theta \approx 20.4°

Jumping the ditch (projectile motion)

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