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A boy can throw a stone up to a maximum height of 10m .The maximum horizontal distance that the boy can throw the same stone up to will be

Option: 1

20 \sqrt{2}m

Option: 2


Option: 3

10 \sqrt{2}m

Option: 4


Answers (1)




Maximum Height-

H= \frac{u^{2}\sin ^{2}\theta }{2g}

Horizontal Range -

Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits.

R=\frac{u^{2}\sin 2\theta }{g}

For maximum horizontal range.

\theta = 45^{0}

R_{max}=\frac{u^{2}\sin 2 (45) }{g}=\frac{u^{2}\times 1}{g}=\frac{u^{2}}{g}


 H=\frac{v^{2_o}sin^2\theta }{2g}  for maximum

                              \Theta =90\ ^{\circ}


given H_{max}=10m

R_{max}= 10=\frac{v^2_0}{2g} -\left ( i \right )

R_{max}= \frac{v^2_0}{2g}= -\left ( ii \right )

\because from equation \left ( i \right ) and \left ( ii \right )

10=\frac{R}{2}\\ \ \ \Rightarrow R=20 meter             

Posted by

Kuldeep Maurya

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