PLEASE I HAVE NO IDEA WHERE TO START
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? Ignore air resistance. Assume the ball is caught at the same height as it is thrown.
Hmm. Sounds lke a "Hail Mary" pass for 98 yards.
The minimum speed is required when the ball is thrown at a 45 degree angle. (As a separate exercise, see if you can prove that). In that case, the ball travels V^2/g horizontally.
Solve the equation
V^2/g = 32.1 m
to obtain the throw velocity, V, required
wow thanks
To determine the minimum speed that the football must have when it leaves the quarterback's hand, we can use the principles of projectile motion. In this case, we need to calculate the initial velocity of the football.
Projectile motion can be divided into horizontal and vertical components. In this scenario, we are only concerned with the horizontal component because we are assuming the ball is thrown at the same height it is caught.
To solve this problem, we can use the equation for horizontal motion:
distance = velocity × time
In this equation, we know the distance (32.1 m) and want to find the minimum velocity. However, since we are concerned with the minimum speed, we can assume that the angle of projection should be 45 degrees. This will give us the maximum horizontal range, which means the minimum speed required.
The horizontal distance traveled can be calculated using the formula:
distance = velocity × time
However, we need to find the time of flight for the given distance. Since the ball travels 32.1 m horizontally, we can assume the time of flight to be the same as it would take for an object to fall vertically from that height.
The formula to calculate the time of flight (t) for an object in free fall is:
distance = initial velocity × time - 0.5 × acceleration × time^2,
where the acceleration is due to gravity (approximately 9.8 m/s^2).
Since the ball is caught at the same height it is thrown, the initial and final heights are the same. So the equation simplifies to:
0 = v₀ × t - 0.5 × 9.8 × t^2,
where v₀ is the initial vertical velocity.
Now we have two equations:
32.1 = v₀ × t, (horizontal motion)
0 = v₀ × t - 0.5 × 9.8 × t^2. (vertical motion)
We can solve the first equation for t:
t = 32.1 / v₀.
Substituting this value of t into the second equation gives:
0 = v₀ × (32.1 / v₀) - 0.5 × 9.8 × (32.1 / v₀)^2.
Simplifying further:
0 = 32.1 - 0.5 × 9.8 × (32.1)^2 / v₀.
Rearranging and multiplying both sides by v₀:
0 = 32.1 × v₀ - 0.5 × 9.8 × (32.1)^2.
Now, we can solve this equation for v₀, which will give us the minimum initial velocity required:
v₀ = (0.5 × 9.8 × (32.1)^2) / 32.1.
Plugging in the values and calculating:
v₀ = (0.5 × 9.8 × (32.1)^2) / 32.1,
v₀ = (0.5 × 9.8 × 1030.41) / 32.1,
v₀ = 5028.189 / 32.1,
v₀ ≈ 156.6 m/s.
So, the minimum speed that the football must have when it leaves the quarterback's hand is approximately 156.6 m/s.