JEE Main Motion in Straight Line PYQ

6. (JEE Main 2022 (Online) 29th July Evening Shift )

A ball is released from a height h. If $t_{1}$ and $t_{2}$ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between $t_{1}$ and $t_{2}$ .

A. $t_{1} = (\sqrt{2}) t_{2}$

B. $t_{1} = (\sqrt{2} - 1) t_{2}$

C. $t_{2} = (\sqrt{2} + 1) t_{1}$

D. $t_{2} = (\sqrt{2} - 1) t_{1}$

Correct Option is (D)

JEE Main 2022 (Online) 29th July Evening Shift Physics - Motion Question 35 English Explanation

For first $\frac{h}{2}$ distance,

$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

$\Rightarrow h = g t_{1}^{2}$

For total distance h,

$h = \frac{1}{2} g (t_{1} + t_{2})^{2}$

$\Rightarrow g t_{1}^{2} = \frac{1}{2} g (t_{1} + t_{2})^{2}$

$\Rightarrow 2 t_{1}^{2} = (t_{1} + t_{2})^{2}$

$\Rightarrow \sqrt{2} t_{1} = t_{1} + t_{2}$

$\Rightarrow (\sqrt{2} - 1) t_{1} = t_{2}$

7. (JEE Main 2022 (Online) 29th July Morning Shift )

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.

A. $\frac{\sqrt{2} - 1}{\sqrt{2} + 1}$

B. $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

C. $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

D. $\frac{1}{3}$

Correct Option is (B)

$v = \sqrt{2 g h}$

$\frac{h}{3} = \sqrt{2 g h} t - \frac{1}{2} g t^{2}$

$\frac{g}{2} t^{2} - \sqrt{2 g h} t + \frac{h}{3} = 0$

$\frac{t_{1}}{t_{2}} = \frac{\sqrt{2 g h} + \sqrt{2 g h - 2 g h / 3}}{\sqrt{2 g h} - \sqrt{2 g h - 2 g h / 3}}$

$= \frac{\sqrt{2} + \frac{2}{\sqrt{3}}}{\sqrt{2} - \frac{2}{\sqrt{3}}} = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

8. (JEE Main 2022 (Online) 28th July Morning Shift )

A NCC parade is going at a uniform speed of $9 km / h$ under a mango tree on which a monkey is sitting at a height of $19.6 m$ . At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is: (Given $g = 9.8 m / s^{2}$ )

A. 5 m

B. 10 m

C. 19.8 m

D. 24.5 m

Correct Option is (A)

$H = \frac{1}{2} g t^{2}$

$19.6 = 4.9 t^{2}$

$t = 2$ sec

$D = 9 \times \frac{5}{18} \times 2 = 5$ m

9. (JEE Main 2022 (Online) 29th June Morning Shift )

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms^{$-$ 1}. [use g = 10 ms^{$-$ 2}] :

A. 10

B. 15

C. 20

D. 30

Correct Option is (D)

As the meeting point lies $100 m$ above ground, displacement of ball will be $80 m$ .

For ball $A$

$\begin{aligned} u = 0, S = 80 m, a = + g = + 10 m / s^{2}, time = t_{1} \\ \Rightarrow S = u t + \frac{1}{2} a t^{2} \\ \Rightarrow 80 = 0 + \frac{1}{2} \times 10 \times t_{1}^{2} \\ \Rightarrow \frac{160}{10} = t_{1}^{2} \\ \Rightarrow t_{1} = 4 s \end{aligned}$

As ball $B$ is thrown after 2 seconds after release of $A$ . Thus, time available for ball $B$ is 2 seconds to cover a distance of $80 m$ .

Let speed be ' $u$ ' $m / s, t_{2} = 4 - 2 = 2 s, S = 80 m$ , $a = + g = + 10 m / s^{2}$

$∴ 80 = u \times 2 + \frac{1}{2} \times 10 \times (2)^{2}$

$\Rightarrow 80 = u + 20$

$\Rightarrow 2 u = 60$

$\Rightarrow u = 30 m / s$

10. .(JEE Main 2022 (Online) 28th July Evening Shift )

A ball is thrown vertically upwards with a velocity of $19.6 {ms}^{- 1}$ from the top of a tower. The ball strikes the ground after $6 s$ . The height from the ground up to which the ball can rise will be $(\frac{k}{5}) m$ . The value of $k$ is __________. (use $g = 9.8 m / s^{2}$ )

Correct answer is (392)

v = 19.6 m/s

t = 6s

Time taken in upward motion above tower = 2s

$\Rightarrow$ Time taken from top most point to ground = 4s

$\Rightarrow \sqrt{\frac{2 h}{g}} = 4$

$h = \frac{16 \times 9.8}{2} = 8 \times 9.8$

$\Rightarrow k = 8 \times 9.8 \times 5 = 392$