A ball is projected from the ground at an angle θ = 45° with speed u = 20 m/s. Find the range and maximum height. (Take g = 10 m/s²)

R = (u² sin 2θ)/(g) = (400 × sin 90°)/(10) = 40 m H = (u² sin²θ)/(2g) = (400 × (1)/(2))/(20) = 10 m

JEE physics

A ball is projected from the ground at an angle θ = 45° with speed u = 20 m/s — JEE Physics

DMDivya Mehta · 10 Asked 1mo ago 1,061 views 1 answer

A ball is projected from the ground at an angle $\theta = 45°$ with speed $u = 20$ m/s. Find the range and maximum height. (Take $g = 10$ m/s²)

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 1mo ago ▲ 6

$$R = \frac{u^2 \sin 2\theta}{g} = \frac{400 \times \sin 90°}{10} = 40 m$$

$$H = \frac{u^2 \sin^2\theta}{2g} = \frac{400 \times \frac{1}{2}}{20} = 10 m$$

Brilliant explanation, the substitution step is what I kept missing.

PrakashGurung52 · 1mo ago

Underrated solution. The way you set it up makes it almost obvious.

DilaniJayawardene31 · 1mo ago

Adding for context: NCERT covers the base concept in the same chapter.

Vikram Nair · 1mo ago

Does this approach generalise to the JEE Advanced version of this question?

Shruti Jain · 1mo ago