At which speed must you throw a beanbag to reach a height of 8 m at a 60° angle?

To determine the speed at which a beanbag must be thrown to reach a height of 8 m at a 60° angle, we can use the principles of projectile motion. The maximum height reached by a projectile can be calculated with the equation:

[ h = \frac{v^2 \sin^2(\theta)}{2g} ]

where ( h ) is the maximum height, ( v ) is the initial velocity, ( \theta ) is the launch angle, and ( g ) is the acceleration due to gravity (approximately ( 9.81 , \text{m/s}^2 )).

Rearranging this formula to solve for ( v ) gives us:

[ v = \sqrt{\frac{2gh}{\sin^2(\theta)}} ]

Substituting ( h = 8 , \text{m} ), ( g = 9.81 , \text{m/s}^2 ), and ( \theta = 60° ) (where ( \sin(60°) = \sqrt{3}/2 )), we can insert the values into the equation:

1. Calculate ( \sin^2(60