A particle is projected from a point on the ground with an initial velocity of u = 100 m/s at angle of 53° with the horizontal
(take, $\tan 53^{\circ} = \frac{4}{3}, g = 10 m / s^{2}$ )

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

The velocity will make an angle of 45° with the horizontal after time 2s .

The average velocity between t = 6s and t =10s is 60m/ s horizontal.

The velocity will make an angle of 45° with the horizontal after time 14s .

Displacement of particle from t = 4to t =12s is 480m.

answer is A, B, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \tan ϕ = \frac{usin gq - gt}{ucos θ} \\ \tan (\pm 45^{\circ}) = \frac{100 \sin 53^{\circ} - 10 t}{100 \cos 53^{\circ}} \\ \Rightarrow t = 2 s and t = 14 s \\ V = ucos θ \hat{i} + (usin θ - gt) \hat{j} \end{array}$
At $t = 6 s, V_{1} = 60 \hat{i} + 20 \hat{j}$
At $t = 10 s, V_{2} = 60 \hat{i} - 20 \hat{j}$
$\begin{array}{l} V_{avg} = \frac{V_{1} + V_{2}}{2} = 60 \hat{i} \\ r = (ucos θ) ti + (usin θt - \frac{1}{2} {gt}^{2}) \hat{j} \end{array}$
At $t = 4 s, r_{1} = 240 \hat{i} + 240 \hat{j}$
At $t = 12 s, r_{2} = 720 \hat{i} + 240 \hat{j}$
Displacement $= |r_{2} - r_{1}| = 480 m$