Question

the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​

Answer 1

Answer:

The actual angle is 30°

Explanation:

Equation of projectile:

y axis:

$v_y(t)=vo*sin(A)-g*t$

the velocity is Zero when the projectile reach in the maximum altitude:

$0=vo-gt\\t=\frac{vo}{g}$

When the time is vo/g the projectile are in the middle of the range.

x axis:

$d_x(t)=vo*cos(A)*t$

R=Range

$R=d_x(t=2*\frac{vo}{g})$

$R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}$

**sin(2A)=2sin(A)cos(A)

The maximum range occurs when A=45°(because sin(90°)=1)

The actual range R'=(2/√3)R:

Let B the actual angle of projectile

$\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}$

2B=60°

B=30°