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1

### JEE Main 2021 (Online) 27th August Evening Shift

A player kicks a football with an initial speed of 25 ms$$-$$1 at an angle of 45$$^\circ$$ from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion ? (Take g = 10 ms$$-$$2)
A
hmax = 10 m

T = 2.5 s
B
hmax = 15.625 m

T = 3.54 s
C
hmax = 15.625 m

T = 1.77 s
D
hmax = 3.54 m

T = 0.125 s

## Explanation

$$H = {{{U^2}{{\sin }^2}\theta } \over {2g}}$$

$$= {{{{(25)}^2}.{{(\sin 45)}^2}} \over {2 \times 10}}$$

= 15.625 m

$$T = {{U\sin \theta } \over g}$$

$$= {{25 \times \sin 45^\circ } \over {10}}$$

= 2.5 $$\times$$ 0.7

= 1.77 s
2

### 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

## Explanation H = $${1 \over 2}$$gt2

$${{9.8 \times 2} \over {9.8}}$$ = t2

t = $$\sqrt 2$$ sec

$$\Delta$$t : time interval between drops

h = $${1 \over 2}$$g($$\sqrt 2$$ $$-$$ 2$$\Delta$$t)2

$$\Delta$$t = $${1 \over {\sqrt 2 }}$$

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

H $$-$$ h = 9.8 $$-$$ 2.45

= 7.35 m
3

### JEE Main 2021 (Online) 26th August Evening Shift

A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a :
A
hyperbola
B
parabola in the direction of motion of plane
C
straight line vertically down the plane
D
parabola in a direction opposite to the motion of plane

## Explanation

Relative velocity of bomb w.r.t. observer in plane = 0.

Bomb will fall down vertically. So, it will move in straight line w.r.t. observer.
4

### 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 $${h \over 3}$$ in both the directions.
A
$${{\sqrt 2 - 1} \over {\sqrt 2 + 1}}$$
B
$${1 \over 3}$$
C
$${{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}$$
D
$${{\sqrt 3 - 1} \over {\sqrt 3 + 1}}$$

## Explanation

$$u = \sqrt {2gh}$$

Now,

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

a = $$-$$g

$$S = ut + {1 \over 2}a{t^2}$$

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

$${t^2}\left( {{g \over 2}} \right) - \sqrt {2gh} t + {h \over 3} = 0$$

$${t_1},{t_2} = {{\sqrt {2gh} \pm \sqrt {2gh - {{4g} \over 2}{h \over 3}} } \over g}$$

$${{{t_1}} \over {{t_2}}} = {{\sqrt {2gh} - \sqrt {{{4gh} \over 3}} } \over {\sqrt {2gh} + \sqrt {{{4gh} \over 3}} }} = {{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}$$

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