A quarterback claims that he can throw the football a horizontal distance of 183 m (200 yd). Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 35° above the horizontal. To evaluate his claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison, a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

Well, let me do some quick calculations here.

First, let's convert the distance from yards to meters. 200 yards is approximately 183 meters, so we'll go with that.

Now, to find the speed at which the quarterback must throw the ball, we need to consider the horizontal and vertical components of the motion.

Since the ball is thrown at an angle of 35° above the horizontal, we can break it down into a horizontal and vertical component. The horizontal component will be the distance the ball travels, which is 183 meters. The vertical component will be the height the ball reaches.

Now, let's focus on the vertical component. The initial vertical velocity, at the point of launch, is zero. The only force acting on the ball vertically is gravity. So, we can use the kinematic equation:

vf^2 = vi^2 + 2ad

Where vf is the final vertical velocity, vi is the initial vertical velocity, a is acceleration, and d is the vertical distance.

Since the final vertical velocity is also zero (at the peak of the throw), we can rewrite the equation as:

0 = 0 + 2ad

Simplifying, we get:

0 = 2ad

Now, we want to find the vertical distance the ball reaches. We can use the equation:

d = (vi*t) + (1/2)*(a*t^2)

Where vi is the initial vertical velocity, t is the time of flight, and a is the acceleration due to gravity (-9.8 m/s^2).

Since we know that d is also zero, we can rewrite the equation as:

0 = (vi*t) + (1/2)*(a*t^2)

Now, let's solve for t:

0 = -4.9*t^2

This equation has two solutions: t = 0 and t = √(0/(-4.9)). Since t cannot be zero (the ball is moving), we have:

t = √(0/(-4.9)) ≈ 0 seconds

Wait a minute... So, the ball would never come down!

Looks like this quarterback's claims are up in the air, quite literally. It seems he would need to throw the ball with an infinite initial speed to achieve that horizontal distance at such a low angle, with no air resistance.

I guess this quarterback might be from another dimension where the laws of physics are a bit different. Would make for an interesting game, that's for sure!

But just for comparison, a baseball pitcher who can accurately throw at 45 m/s (100 mph) is truly exceptional. So, let's stick to that reality for now.

To determine the speed with which the quarterback must throw the ball, we can use the following projectile motion equation:

R = (V² * sin(2θ)) / g

where:
R = horizontal distance (183 m),
V = initial velocity of the throw,
θ = launch angle (35°),
g = acceleration due to gravity (9.8 m/s²).

Using this equation, we can solve for V.

Substituting the given values into the equation, we have:

183 = (V² * sin(2 * 35°)) / 9.8

To solve for V, we need to isolate it on one side of the equation. Multiply both sides of the equation by 9.8:

183 * 9.8 = V² * sin(2 * 35°)

Simplifying further:

1793.4 = V² * sin(70°)

Next, divide both sides of the equation by sin(70°):

1793.4 / sin(70°) = V²

Finally, take the square root of both sides to solve for V:

V = √(1793.4 / sin(70°))

Using a calculator, we find that:

V ≈ 52.9 m/s

Therefore, the quarterback must throw the ball at a speed of approximately 52.9 m/s to achieve a horizontal distance of 183 m.

Note: Comparing this speed to a baseball pitcher's fastball of 45 m/s (100 mph), the quarterback's throwing speed is exceptional.

To determine the speed with which the quarterback must throw the ball, we can use the principles of projectile motion.

The horizontal and vertical motions of the football can be treated independently.

First, let's consider the vertical motion. The initial vertical velocity of the football can be determined using the formula:

v_initial_vertical = v_initial * sin(theta)

where v_initial is the initial velocity of the ball and theta is the launch angle (35° in this case).

Next, we can determine the time it takes for the ball to reach the peak of its trajectory (where the vertical velocity becomes zero). At the peak, the vertical displacement is equal to zero. We can use the formula:

v_final_vertical = v_initial_vertical + (g * t)

where v_final_vertical is the final vertical velocity of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken to reach the peak.

Since the vertical displacement is zero at the peak, we can rearrange the formula to solve for t:

0 = v_initial_vertical + (g * t)

Solving for t gives us:

t = -v_initial_vertical / g

Once we have the time taken to reach the peak, we can use this to calculate the total time of flight of the football, since the total time of flight is twice the time taken to reach the peak.

t_total = 2 * t

Now, let's focus on the horizontal motion. The horizontal displacement of the ball can be calculated using the formula:

horizontal_displacement = v_horizontal * t_total

where v_horizontal is the horizontal velocity of the ball.

Since the horizontal displacement is given as 183 m, we can rearrange the formula to solve for v_horizontal:

v_horizontal = horizontal_displacement / t_total

Now we can calculate the required speed:

speed = sqrt(v_horizontal^2 + v_initial_vertical^2)

Substituting the values we have:

horizontal_displacement = 183 m
theta = 35°
g = 9.8 m/s^2

We can now plug these values into the equations to calculate the required speed.