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A quarterback claims that he can throw the football a horizontal distance of 196.6 m (215 yd). Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 30° above the horizontal. To evaluate his claim, determine the speed with which this quarterback must throw the ball.
Given a kickoff of velocity v at angle a, the range is given by
R = v^2 sin(2a)/g
196.6 = v^2 * .866 / 9.8
v^2 = 2224.8
v = 47.17m/s
v =~ 170km/hr, not too likely, especially since no quarterback would be discussing distances in meters!
To determine the speed with which the quarterback must throw the ball, we can use the principles of projectile motion.
Let's break down the problem into its key components:
1. The horizontal distance covered by the ball (range) is 196.6 m.
2. The ball is launched at an angle of 30° above the horizontal.
To solve this problem, we will use the following kinematic equation for the range of projectile motion:
Range = (initial velocity)^2 * sin(2θ) / g
Where:
- Range is the horizontal distance covered by the ball
- θ is the launch angle (30°)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Rearranging the equation, we can solve for the initial velocity:
(initial velocity)^2 = Range * g / sin(2θ)
Now, we can plug in the given values and calculate the initial velocity:
Range = 196.6 m
θ = 30°
g = 9.8 m/s^2
(initial velocity)^2 = 196.6 * 9.8 / sin(2 * 30°)
To find sin(60°), we can use the trigonometric identity sin(2θ) = 2 * sin(θ) * cos(θ):
(initial velocity)^2 = 196.6 * 9.8 / (2 * sin(30°) * cos(30°))
(sin(30°) = 0.5 and cos(30°) = √(3)/2)
(initial velocity)^2 = 196.6 * 9.8 / (2 * 0.5 * √(3)/2)
(initial velocity)^2 = 196.6 * 9.8 / (√(3))
(initial velocity)^2 = 3422.932
Taking the square root of both sides, we can find the initial velocity:
initial velocity = √(3422.932)
Therefore, the speed with which the quarterback must throw the ball is approximately 58.55 m/s.