Question

Two projectiles A and B are projected with same speed at angles ${30}^0$ and ${60}^0$ to horizontal, then choose the wrong statement? (symbols have their usual meaning)

• A. $R_A=R_B$
• B. $H_B=3H_A$
• C. $\sqrt{3}{T}_B=T_A$
• D. All of these

Answer 1

Answer: (A)

$R_A=R_B$

$Given:$ Two projectiles A and B are projected with same speed at angles ${30}^o$ and ${60}^o$ to horizontal.

$Solution:$ In projectile motion the time of flight, maximum height and range is given by $T=\frac{2u\sin \theta }{g},H=\frac{u^2{\sin }^2\theta }{2g}$ and $R=\frac{u^2\sin 2\theta }{g}$

for projectile a $\theta ={30}^{\circ }$ so $T_a=\frac{2u\sin {30}^{\circ }}{g}=\frac{1}{2}\cdot \frac{2u}{g}$

$H_a=\frac{u^2{\sin }^2{30}^{\circ }}{2g}=\frac{1}{4}\frac{u^2}{2g}$ and

$R_a=\frac{u^2\sin {60}^{\circ }}{g}=\frac{\sqrt{3}}{2}\frac{u^2}{g}$

for projectile b $\theta ={60}^{\circ }$ so $T_b=\frac{2u\sin {60}^{\circ }}{g}=\frac{\sqrt{3}}{2}\cdot \frac{2u}{g}$

$H_b=\frac{u^2{\sin }^2{60}^{\circ }}{2g}=\frac{3}{4}\frac{u^2}{2g}$ and

$R_a=\frac{u^2\sin {120}^{\circ }}{g}=\frac{\sqrt{3}}{2}\frac{u^2}{g}$

Hence $T_b=\sqrt{3}\cdot T_a$

$H_b=3\cdot T_a$

$R_b=R_a$

$So,the$ $correct$ $option:A$