From the top of a tower a ball is thrown vertically upwards which reaches the ground in 6s

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)

Correct Answer is 20

$$ \Rightarrow v\cos \alpha = u\cos 45^\circ $$ ..... (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

एक पिण्ड क्षैतिज से $$45^{\circ}$$ के कोण पर प्रक्षेपित किया जाता है । इसका वेग $$2 \mathrm{~s}$$ बाद $$20 \mathrm{~ms}^{-1}$$ है | गति के दौरान, पिण्ड द्वारा प्राप्त अधिकतम ऊँचाई का मान _____________ $$\mathrm{m}$$ होगा । (यदि $$\mathrm{g}$$ $$=10 \mathrm{~ms}^{-2}$$ )

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