Question

The range of a projectile that is launched with an initial velocity v at an angle of a with the horizontal is given by R

sin

where g is the acceleration due to gravity or 9.8 meters per second squared. If a projectile is launched with an initial velocity of 1

meters per second, what angle is required to achieve a range of 20 meters? Round answers to the nearest whole number.

Answer 1

`\theta=30.285^(\circ)`

Explanation:

The range of a projectile is given by :

`R=(u^2\sin2\theta)/(g)`

Put R = 20 m, u = 15 m/s and finding the value of angle of projection

So,

`R=(u^2\sin2\theta)/(g)\\\\\sin2\theta=(Rg)/(u^2)\\\\\sin2\theta=(20* 9.8)/(15^2)\\\\\sin2\theta=0.871\\\\2\theta=\sin^(-1)(0.871)\\\\2\theta=60.57\\\\\theta=30.285^(\circ)`

So, the required angle of projection is equal to
`30.285^(\circ)`.