A quarterback is asked to throw a football to a receiver that is 32.1 m away. What is the minimum speed that the football must have when it leaves the quarterback's hand?
The minimum speed required for a given distance is achieved when the ball is thrown at a 45 degree angle. In that case,
X = Vo^2/g
For other launch angles, the horizontal distance covered is
(Vo^2/g) * sin 2A
where A is the launch angle, measured from horizontal.
Use the first equation to solve for the initial velocity Vo.
To find the minimum speed that the football must have when it leaves the quarterback's hand, you can use the principle of conservation of energy.
The initial energy of the football is purely kinetic energy, given by the equation:
Kinetic Energy = (1/2) * mass * velocity^2
Where mass is the mass of the football and velocity is the speed at which it leaves the quarterback's hand.
The final energy of the football is a combination of kinetic energy and potential energy, given by the equation:
Final Energy = (1/2) * mass * velocity^2 + mass * g * height
Where g is the acceleration due to gravity (approximately 9.8 m/s^2) and height is the vertical distance between the quarterback and receiver.
Since energy is conserved, the initial energy must be equal to the final energy. As the football is thrown horizontally, there is no change in height (height = 0), so the equation becomes:
(1/2) * mass * velocity^2 = (1/2) * mass * velocity^2 + mass * g * 0
Simplifying the equation, we get:
(1/2) * mass * velocity^2 = (1/2) * mass * velocity^2
The mass of the football cancels out from both sides of the equation, leaving us with:
velocity^2 = velocity^2
Taking the square root of both sides, we find that:
velocity = velocity
Therefore, any non-zero velocity is acceptable for the minimum speed that the football must have when it leaves the quarterback's hand to reach a receiver 32.1 m away.