JEE Main Motion in Straight Line PYQ

21. (JEE Main 2014 (Offline) )

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

A. 2gH = n²u²

B. gH = (n - 2)²u²

C. 2gH = nu²(n - 2)

D. gH = (n - 2)u²

Correct Option is (C)

JEE Main 2014 (Offline) Physics - Motion Question 142 English Explanation

Time taken to reach highest point is $t = \frac{u}{g}$

Time taken by the particle to reach the ground = $n t = \frac{n u}{g}$

Speed on reaching ground $v = \sqrt{u^{2} + 2 g H}$

Now, $v = u + a t$

$\Rightarrow \sqrt{u^{2} + 2 g H} = - u + g t$

$\Rightarrow t = \frac{u + \sqrt{u^{2} + 2 g H}}{g} = \frac{n u}{g}$ (from question)

$\Rightarrow 2 g H = n (n - 2) u^{2}$

22. (AIEEE 2005 )

A parachutist after bailing out falls $50$ $m$ without friction. When parachute opens, it decelerates at $2 m / s^{2} .$ He reaches the ground with a speed of $3$ $m / s$ . At what height, did he bail out?

A. $182$ $m$

B. $91$ $m$

C. $111$ $m$

D. $293$ $m$

Correct Option is (D)

AIEEE 2005 Physics - Motion Question 140 English Explanation
The velocity of parachutist when parachute opens at 50 m is

$u = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 50} = \sqrt{980}$

The velocity at ground, $v = 3 m / s$

$∴$ $S = \frac{v^{2} - u^{2}}{2 \times (- 2)} = \frac{3^{2} - 980}{- 4} \approx 243 m$

Initially he has fallen $50$ $m .$

$∴$ Total height from where

He bailed out $= 243 + 50 = 293 m$

23. (AIEEE 2004 )

A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in $\frac{T}{3}$ seconds?

A. $\frac{8 h}{9}$ meters from the ground

B. $\frac{7 h}{9}$ meters from the ground

C. $\frac{h}{9}$ meters from the ground

D. $\frac{7 h}{18}$ meters from the ground

Correct Option is (A)

We know that equation of motion, $s = u t + \frac{1}{2} g t^{2},$

Initial speed of ball is zero and it take T second to reach the ground.

$∴$ $h = \frac{1}{2} g T^{2}$

After $T / 3$ second, vertical distance moved by the ball

$h^{'} = \frac{1}{2} g {(\frac{T}{3})}^{2}$

$\Rightarrow h^{'} = \frac{1}{2} \times \frac{8 T^{2}}{9}$

$= \frac{h}{9}$

$∴$ Height from ground

$= h - \frac{h}{9} = \frac{8 h}{9}$