JEE Main Motion in Straight Line PYQ

16. (JEE Main 2021 (Online) 27th August Evening Shift )

Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.

A. 4.18 m

B. 2.94 m

C. 2.45 m

D. 7.35 m

Correct Option is (D)

JEE Main 2021 (Online) 27th August Evening Shift Physics - Motion Question 72 English Explanation
H = $\frac{1}{2}$ gt²

$\frac{9.8 \times 2}{9.8}$ = t²

t = $\sqrt{2}$ sec

$Δ$ t : time interval between drops

h = $\frac{1}{2}$ g( $\sqrt{2}$ $-$ 2 $Δ$ t)²

$Δ$ t = $\frac{1}{\sqrt{2}}$

h = $\frac{1}{2}$ g ${(\sqrt{2} - \frac{1}{\sqrt{2}})}^{2} = \frac{1}{2} \times 9.8 \times \frac{1}{2} = \frac{9.8}{4} = 2.45$ m

H $-$ h = 9.8 $-$ 2.45

= 7.35 m

17. (JEE Main 2021 (Online) 27th July Morning Shift )

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching $\frac{h}{3}$ in both the directions.

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

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

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

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

Correct Option is (C)

$u = \sqrt{2 g h}$

Now,

$S = \frac{h}{3}$

a = $-$ g

$S = u t + \frac{1}{2} a t^{2}$

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

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

From quadratic equation

$t_{1}, t_{2} = \frac{\sqrt{2 g h} \pm \sqrt{2 g h - \frac{4 g}{2} \frac{h}{3}}}{g}$

$\frac{t_{1}}{t_{2}} = \frac{\sqrt{2 g h} - \sqrt{\frac{4 g h}{3}}}{\sqrt{2 g h} + \sqrt{\frac{4 g h}{3}}} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

18. (JEE Main 2021 (Online) 25th July Morning Shift )

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4^th second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap ? (Take g = 9.8 m/s²)

A. 3 drops / 2 sconds

B. 2 drops / second

C. 1 drop / second

D. 1 drop / 7 seconds

Correct Option is (C)

In 4 sec. 1^st drop will travel

$\Rightarrow \frac{1}{2}$ $\times$ (9.8) $\times$ (4)² = 78.4 m

$∴$ 2^nd drop would have travelled

$\Rightarrow$ 78.4 $-$ 34.3 = 44.1 m.

Time for 2^nd drop

$\Rightarrow \frac{1}{2}$ (9.8)t² = 44.1

t = 3 sec

$∴$ each drop have time gap of 1 sec

$∴$ 1 drop per sec

19. .(JEE Main 2021 (Online) 26th August Morning Shift )

Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3s with the same initial velocity of 35 m/s, then these balls collide at a height of ............... m. (Take g = 10 m/s²)

Correct answer is (50)

JEE Main 2021 (Online) 26th August Morning Shift Physics - Motion Question 76 English Explanation

When both balls will collide

y₁ = y₂

$35 t - \frac{1}{2} \times 10 \times t^{2} = 35 (t - 3) - \frac{1}{2} \times 10 \times (t - 3)^{2}$

$35 t - \frac{1}{2} \times 10 \times t^{2} = 35 t - 105 - \frac{1}{2} \times 10 \times t^{2} - \frac{1}{2} \times 10 \times 3^{2} + \frac{1}{2} \times 10 \times 6 t$

0 = 150 $-$ 30 t

t = 5 sec

$∴$ Height at which both balls will collied

$h = 35 t - \frac{1}{2} \times 10 \times t^{2}$

$= 35 \times 5 - \frac{1}{2} \times 10 \times 5^{2}$

h = 50 m

20. (JEE Main 2020 (Online) 5th September Morning Slot )

A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to :
[g is the acceleration due to gravity]

A. t = 3.4 $\sqrt{(\frac{h}{g})}$

B. t = 1.8 $\sqrt{(\frac{h}{g})}$

C. t = $\sqrt{\frac{2 h}{3 g}}$

D. t = $\frac{2}{3} \sqrt{(\frac{h}{g})}$

Correct Option is (A)

JEE Main 2020 (Online) 5th September Morning Slot Physics - Motion Question 103 English Explanation
Velocity of helicopter at height h,

$V_{B}^{2} = 0^{2} + 2 g h$

$V_{B} = \sqrt{2 g h}$

$- h = (V_{B}) t - \frac{1}{2} g t^{2}$

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

$\Rightarrow$ $g t^{2} - 2 \sqrt{2 g h t} - 2 h = 0$

$\Rightarrow$ $t = \frac{\sqrt{2 g h t} \pm \sqrt{8 g h + 8 g h}}{2 g} = \frac{2 \sqrt{2 g h} \pm \sqrt{16 g h}}{2 g} = \frac{\sqrt{2 g h} + 2 \sqrt{g h}}{g}$

$\Rightarrow$ $t = \sqrt{\frac{2 h}{g}} + 2 \sqrt{\frac{h}{g}} = \sqrt{\frac{h}{g}} (\sqrt{2} + 2) = 3.4 \sqrt{\frac{h}{g}}$