1
JEE Main 2022 (Online) 27th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A projectile is launched at an angle '$$\alpha$$' with the horizontal with a velocity 20 ms$$-$$1. After 10 s, its inclination with horizontal is '$$\beta$$'. The value of tan$$\beta$$ will be : (g = 10 ms$$-$$2).

A
tan$$\alpha$$ + 5sec$$\alpha$$
B
tan$$\alpha$$ $$-$$ 5sec$$\alpha$$
C
2tan$$\alpha$$ $$-$$ 5sec$$\alpha$$
D
2tan$$\alpha$$ $$+$$ 5sec$$\alpha$$
2
JEE Main 2022 (Online) 27th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A girl standing on road holds her umbrella at 45$$^\circ$$ with the vertical to keep the rain away. If she starts running without umbrella with a speed of 15$$\sqrt2$$ kmh$$-$$1, the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is :

A
30 kmh$$-$$1
B
$${{25} \over {\sqrt 2 }}$$ kmh$$-$$1
C
$${{30} \over {\sqrt 2 }}$$ kmh$$-$$1
D
25 kmh$$-$$1
3
JEE Main 2022 (Online) 25th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. IF A and B reached the maximum height h1 and h2 respectively, then $$R = 4\sqrt {{h_1}{h_2}} $$

Reason R : Product of said heights.

$${h_1}{h_2} = \left( {{{{u^2}{{\sin }^2}\theta } \over {2g}}} \right)\,.\,\left( {{{{u^2}{{\cos }^2}\theta } \over {2g}}} \right)$$

Choose the correct answer :

A
Both A and R are true and R is the correct explanation of A.
B
Both A and R are true but R is NOT the correct explanation of A.
C
A is true but R is false.
D
A is false but R is true.
4
JEE Main 2022 (Online) 24th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A projectile is projected with velocity of 25 m/s at an angle $$\theta$$ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of $$\theta$$ will be :

[use g = 10 m/s2]

A
$${1 \over 2}{\sin ^{ - 1}}\left( {{{5{t^2}} \over {4R}}} \right)$$
B
$${1 \over 2}{\sin ^{ - 1}}\left( {{{4R} \over {5{t^2}}}} \right)$$
C
$${\tan ^{ - 1}}\left( {{{4{t^2}} \over {5R}}} \right)$$
D
$${\cot ^{ - 1}}\left( {{R \over {20{t^2}}}} \right)$$
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