JEE MAIN - Physics (2022 - 24th June Evening Shift - No. 20)
A body is projected from the ground at an angle of 45$$^\circ$$ with the horizontal. Its velocity after 2s is 20 ms$$-$$1. The maximum height reached by the body during its motion is __________ m. (use g = 10 ms$$-$$2)
Answer
20
Explanation
_24th_June_Evening_Shift_en_20_1.png)
$$ \Rightarrow v\cos \alpha = u\cos 45^\circ $$ ..... (i)
& $$v\sin \alpha = u\sin 45^\circ - gt$$ ..... (ii)
Solve for u we get
$$u = 20\sqrt 2 $$ m/s
$$ \Rightarrow H = {{{u^2}{{\sin }^2}45^\circ } \over {20}} = 20$$ m
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