An arrow is projected into air. Its time of flight is 5 ~s and range 200 ~m . What is the maximum height reached by it? (Take g=10 ~m ~s^ -2 )

Correct answer: A. 31.25 ~m Given, 5=(2 u θ)/(g) or (u θ)/(g)=(5)/(2) Maximum height =(u^2 ^2 θ)/(2 g)=(g)/(2) ×((u^2 ^2 θ)/(g^2)) =(g)/(2) ×((5)/(2))^2=(10)/(2) × (25)/(4)=31.25 ~m

AIIMS 2014

Exam: AIIMS
Year: 2014
Subject: Physics
Chapter: Motion in Two Dimensions

AIIMS 2014 · Physics · Motion in Two Dimensions

An arrow is projected into air. Its time of flight is 5 ~s and...

An arrow is projected into air. Its time of flight is

5 \mathrm{~s}

and range

200 \mathrm{~m}

. What is the maximum height reached by it? (Take

g=10 \mathrm{~m} \mathrm{~s}^{-2}

)

Options

A $31.25 \mathrm{~m}$
B $24.5 \mathrm{~m}$
C $18.25 \mathrm{~m}$
D $46.75 \mathrm{~m}$

Correct answer

A. $31.25 \mathrm{~m}$

Step-by-step solution

Given,

5=\frac{2 u \sin \theta}{g} or

\frac{u \sin \theta}{g}=\frac{5}{2}

Maximum height

=\frac{u^2 \sin ^2 \theta}{2 g}=\frac{g}{2} \times\left(\frac{u^2 \sin ^2 \theta}{g^2}\right)

=\frac{g}{2} \times\left(\frac{5}{2}\right)^2=\frac{10}{2} \times \frac{25}{4}=31.25 \mathrm{~m}

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