Question

A stone is projected with a velocity u at angle $\theta$ with the horizontal reaches maximum height H1, when it is projected with a velocity u at an angle $\left(\frac{\pi }{2}-\theta \right)$ with the horizontal, it reaches to a maximum height H2. The relation between the horizontal range R of the projectile, H1 and H2 is:

• A. $R=4\sqrt{H_1{H}_2}$
• B. $R=4(H_1-H_2)$
• C. $R=4(H_1+H_2)$
• D. $R={H_1}^2/{H_2}^2$

Answer 1

Answer: (A)

$H_1=\frac{u^{2}si{n}^2\theta }{2g}$ $H_2=\frac{u^2{\sin }^2(90-\theta )}{2g}=\frac{u^2{\cos }^{2}2\theta }{2g}$ From $\frac{P_1{V}_1}{T_1}=\frac{T_2{V}_2}{T_2}$ $\frac{P_1{m}_1}{\rho T_1}=\frac{P_2{m}_2}{\rho T_2}$ $\frac{720\times m}{313}=\frac{P_2\left(m-\frac{1}{4}m\right)}{626}$ $\frac{720\times m}{313}=\frac{P_2\times \frac{3}{4}m}{626}$ $P_2=\frac{720\times 626\times 4}{313\times 3}$ $=1920kPa.$