Projectile Motion: Diagram-Based Questions (3)

hard • 2 min read
Tags Projectile Motion

Question

A projectile is fired from the top of a 2020 m cliff at u=30u = 30 m/s at angle θ=30°\theta = 30° above the horizontal. Find (a) time of flight, and (b) horizontal range from the launch point. Take g=10g = 10 m/s2^2.

Solution — Step by Step

ux=ucos⁡θ=30×32≈25.98 m/su_x = u \cos\theta = 30 \times \frac{\sqrt{3}}{2} \approx 25.98 \text{ m/s}

uy=usin⁡θ=30×12=15 m/su_y = u \sin\theta = 30 \times \frac{1}{2} = 15 \text{ m/s}

Take upward as positive; initial height h=20h = 20 m above the landing point. The vertical equation:

−h=uyt−12gt2-h = u_y t - \frac{1}{2} g t^2

−20=15t−5t2  ⟹  5t2−15t−20=0-20 = 15 t - 5 t^2 \implies 5 t^2 - 15 t - 20 = 0

t2−3t−4=0  ⟹  (t−4)(t+1)=0t^2 - 3t - 4 = 0 \implies (t-4)(t+1) = 0

So t=4t = 4 s (rejecting negative root).

R=ux⋅t=25.98×4≈103.9 mR = u_x \cdot t = 25.98 \times 4 \approx 103.9 \text{ m}

Final answers: T=4T = 4 s, R≈103.9R \approx 103.9 m.

Why This Works

Horizontal motion is uniform (no horizontal force after launch); vertical motion is uniformly accelerated under gravity. Treat them independently and link via the same time variable. The cliff makes the vertical equation a quadratic in tt — two roots, take the positive one.

For a level-ground projectile, the formula T=2usin⁡θ/gT = 2u\sin\theta/g works. From a cliff, you must solve the full quadratic — students sometimes wrongly apply the level-ground formula and get a smaller time.

Alternative Method

Energy + momentum: at landing, vertical speed vyv_y satisfies vy2=uy2+2gh=225+400=625v_y^2 = u_y^2 + 2 g h = 225 + 400 = 625, so vy=25v_y = 25 m/s. Then T=(vy−uy)/g⋅T = (v_y - u_y)/g \cdot … actually it’s (vy+uy)/g=(25+15)/10=4(v_y + u_y)/g = (25 + 15)/10 = 4 s. Cross-check: matches the kinematics result.

For projectiles from a height, always set up the coordinate system carefully and write the displacement equation directly. Avoid the level-ground shortcut formulas — they don’t apply.

Common Mistake

Students apply T=2usin⁡θ/g=3T = 2u\sin\theta/g = 3 s, treating it like level ground. That gives the wrong range. From an elevated launch, the projectile spends extra time falling below the launch level — solve the full quadratic.

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