A projectile is thrown from ground towards a large vertical wall which is at a distance of 8 m from the point of projection. The launching speed of projectile from ground is $102 m / s$ If projectile hits the wall at a height h from ground, then maximum possible value of h is ……..(in meter)
(Take $g = 10 {ms}^{- 2}$ )

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 8.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$y = x \tan θ - \frac{g x^{2}}{2 u^{2}} \sec^{2} θ$

$h = L \tan θ - \frac{g L^{2}}{2 u^{2}} (1 + \tan^{2} θ)$
$For h to be maximum, \frac{dh}{dθ} = 0$

$\Rightarrow u^{2} = L g \tan θ$

$∴ \tan θ = \frac{{(10 \sqrt{2})}^{2}}{8 (10)} = \frac{5}{2}$
$h_{\max} = 8 (\frac{5}{2}) - \frac{10}{2} \frac{8^{2}}{200} (\frac{29}{4}) = 8.4 m$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE