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.
Multiply by gravity and take the square root
To find the minimum speed that the football must have when it leaves the quarterback's hand, you can use the principles of projectile motion. Since the given distance is horizontal, we can use the formula for horizontal distance, which is:
distance (d) = initial velocity (v) * time (t)
Since the quarterback throws the ball at an angle, we need to analyze the vertical and horizontal components of the motion separately. In the vertical direction, the ball experiences free fall, so we can use the equation:
vertical distance (h) = vertical initial velocity (v_y) * time (t) + 0.5 * acceleration due to gravity (g) * time (t)^2
where the vertical initial velocity is zero because the ball is thrown horizontally.
Since we want to find the minimum speed, we are looking for the minimum time it takes for the ball to reach the receiver. At the minimum time, the vertical distance h will be equal to zero because the ball will be at the same height when thrown and caught.
Setting h = 0 and rearranging the equation, we get:
0 = 0 + 0.5 * g * t^2
Simplifying the equation, we have:
0.5 * g * t^2 = 0
Since time t cannot be zero, the only solution is for acceleration due to gravity (g) to be zero.
Therefore, the ball should be thrown horizontally to minimize the time it takes to reach the receiver. In this case, the minimum speed required is determined by the horizontal distance, which is 32.1 meters.
Using the horizontal distance formula mentioned earlier, we can rearrange the equation to solve for initial velocity (v):
v = d / t
Since the ball is thrown horizontally, the time it takes to travel the distance is equal to the time of flight (t). Therefore, the minimum speed the ball must have is:
v = 32.1 m / t
To find the value of time (t), we can calculate it using the horizontal distance and the horizontal component of the ball's velocity. Since the vertical initial velocity v_y is zero, we can calculate the time as:
t = d / v_x
where v_x is the horizontal component of the ball's velocity.
Since the ball is thrown at an angle, we need to find the horizontal component of the velocity. We can do this by using trigonometric functions. If θ represents the angle from the horizontal axis at which the ball is thrown, we have:
cos(θ) = v_x / v
Rearranging the equation, we get:
v_x = v * cos(θ)
Now we can substitute this expression for v_x into the equation for time:
t = d / (v * cos(θ))
Finally, substituting this expression for time (t) back into the equation for the minimum speed (v):
v = 32.1 m / (d / (v * cos(θ)))
Simplifying further:
v = v * cos(θ) * (32.1 m / d)
Now we can solve for the minimum speed:
v = (v * cos(θ) * 32.1 m) / d
To find the value of the minimum speed, we need to cancel out the variable v. We can do this by rearranging the equation:
v / v = (32.1 m * cos(θ)) / d
1 = (32.1 m * cos(θ)) / d
Multiplying both sides by d:
d = 32.1 m * cos(θ)
Finally, rearranging the equation to solve for v:
v = d / cos(θ)
Now we can substitute the given horizontal distance (d) of 32.1 m and assume an angle (θ) at which the ball is thrown, and calculate the minimum speed using the equation:
v = 32.1 m / cos(θ)