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

The actual angle is 30°

Step-by-step 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=(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*(vo)/(g))`

`R=vo*cos(A)*2(vo)/(g) \\\\R=((vo)^(2)*2* sin(A)cos(A))/(g) \\\\R=((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

`(vo^(2) )/(g) =((2)/(√(3) )) (vo^(2) *sin(2B))/(g)\\\\1= (2 )/(√(3)) *sin(2B)\\\\sin(2B)=(√(3))/(2)\\\\`

2B=60°

B=30°