Kinematics

JEE Main level, single correct, medium

282 of 308

Question 282 of 308

A boy playing on the roof of a 10 m10 \mathrm{~m} high building throws a ball with a speed of 10 m/s10 \mathrm{~m} / \mathrm{s} at an angle of 30∘30^{\circ} with the horizontal. How far from the throwing point will the ball be at the height of 10 m10 \mathrm{~m} from the ground?
[g=10 m/s2,sin⁡30∘=12,cos⁡30∘=32]\left[\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sin 30^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]

Show answer and solution

Answer: (D) 8.66 m8.66 \mathrm{~m}

From, the question if the horizontal distance is none other than the horizontal range on the level of the roof of building
 Range =u2sin⁡2θg=(10)2sin⁡(2×30)g=10×10×32×10=8.66\text { Range }=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}=\frac{(10)^2 \sin (2 \times 30)}{\mathrm{g}}=\frac{10 \times 10 \times \sqrt{3}}{2 \times 10}=8.66

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Projectile motion simulationDrag the sliders and watch the numbers in this question change.Revise Kinematics & Projectile Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Kinematics (Projectile motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.