Question

For a projectile thrown into space with a speed v, the horizontal range is $\frac{\sqrt{3}v^{2}}{2g}$. The vertical range is $\frac{v^{2}}{8g}$. The angle which the projectile makes with the horizontal initially is-

• A. 15⁰
• B. 30⁰
• C. 45⁰
• D. 60⁰

Answer 1

Answer: (B)

30⁰

Answer 2

$\frac { u ^ { 2 } \sin 2 \theta } { g }$ = $\frac { \sqrt { 3 } v ^ { 2 } } { 2 g }$ ̃ sin2q = $\frac { \sqrt { 3 } } { 2 }$ ̃ q = 30⁰

$\frac { u ^ { 2 } \sin ^ { 2 } \theta } { 2 g }$ = $\frac { \mathrm { u } ^ { 2 } } { 8 \mathrm {~g} }$ ̃ sinq = $\frac { 1 } { 2 }$ or q = 30⁰