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

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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

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
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