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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
$10m$

Option: 3
$10 \sqrt{2}m$

Option: 4
$20m$

Answers (1)

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Solution

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

$\Theta =90\ ^{\circ}$

$H_{max}=\frac{v^2_0}{2g}$

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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