Question

If R is the horizontal range for inclination and H is the height reached by the projectile show that R(max.)is given by Rmax=4h

Answer 1

Explanation:

R=

${u}^{2} \sin(2 \alpha ) \div g$

For R to be max 2à must be 90 i.e

$\alpha = {45}^{o}$

i.e. R=

${u}^{2} \div g$

And H=

${u}^{2} { \sin(45) }^{2} \div 2g$

=

${u }^{2} \div 4g$

Since, R max=

${u}^{2} \div g$

So, Rmax=4h

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