Kinematics: Step-by-Step Worked Examples (6)

hard 3 min read

Question

A particle is projected from the ground with speed u=20u = 20 m/s at an angle of 6060^\circ above horizontal. A wall stands 2020 m away. Find the height at which the particle strikes the wall and whether it is still rising or falling at that instant. Take g=10g = 10 m/s2^2.

Solution — Step by Step

Horizontal: ux=ucos60=20×0.5=10u_x = u\cos 60^\circ = 20 \times 0.5 = 10 m/s. Vertical: uy=usin60=20×32=103u_y = u\sin 60^\circ = 20 \times \tfrac{\sqrt{3}}{2} = 10\sqrt{3} m/s. The horizontal motion is uniform; the vertical motion has acceleration g-g.

The wall is x=20x = 20 m away, so t=xux=2010=2t = \dfrac{x}{u_x} = \dfrac{20}{10} = 2 s. We use horizontal motion because that’s the only equation where time and distance link cleanly.

y=uyt12gt2=103(2)12(10)(4)=2032014.64 my = u_y t - \tfrac{1}{2} g t^2 = 10\sqrt{3}(2) - \tfrac{1}{2}(10)(4) = 20\sqrt{3} - 20 \approx 14.64 \text{ m}

So the particle hits the wall at roughly 14.64 m above the ground.

Vertical velocity at t=2t = 2 s is vy=uygt=103202.68v_y = u_y - g t = 10\sqrt{3} - 20 \approx -2.68 m/s. The negative sign tells us it is falling when it strikes the wall.

Why This Works

Projectile motion always splits into two independent 1D problems — uniform horizontal and uniformly accelerated vertical. The link between them is time. Once we know tt from horizontal motion, we plug it into the vertical equation and the rest is arithmetic.

Students sometimes try to use the trajectory equation y=xtanθgx22u2cos2θy = x\tan\theta - \dfrac{g x^2}{2u^2\cos^2\theta} directly. That works, but for finding the velocity direction at impact, the time-based method is faster and more intuitive.

Alternative Method

Using the trajectory equation:

y=20tan6010(20)22(20)2cos260=2034000200=2032014.64 my = 20\tan 60^\circ - \frac{10(20)^2}{2(20)^2\cos^2 60^\circ} = 20\sqrt{3} - \frac{4000}{200} = 20\sqrt{3} - 20 \approx 14.64 \text{ m}

Same answer. Pick the method you remember under exam pressure.

Many students forget that sin60=32\sin 60^\circ = \dfrac{\sqrt{3}}{2}, not 12\dfrac{1}{2}. Mixing up sine and cosine of 6060^\circ is the single most common JEE Main projectile error — it changes the answer drastically.

If the question asks “rising or falling,” compute vyv_y at that instant. Sign of vyv_y tells you immediately. No need to find peak time.

Final answer: height 14.64\approx 14.64 m, particle is falling.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next