A football is thrown to a moving
football player. The football leaves the quarterback’s
hands 1.6m above the ground with a speed of 14 m/s
at an angle 30o above the horizontal. If the receiver
starts 11 m away from the quarterback along the line
of flight of the ball when it is thrown, what constant
velocity must he have to get to the ball at the instant it
is 1.6m above the ground?
At time t, the ball's height is
h(t) = 1.6 + (14sin30°)t - 4.9t^2
So, the time it takes for the ball to go up and drop back to 1.6 is 1.428 seconds.
The ball's constant horizontal speed is 14cos30° = 12.124 m/s
So, the ball travels 12.124*1.428 = 17.314m
During that time t, the runner's distance is 11+1.428v. So, find v such that
11 + 1.428v = 17.314
v = 4.42 m/s
Since the final height is the same as the initial height call them both 0, not 1.6 to make it easy.
Vi = initial speed up = 14 * sin 30 = 7 m/s upward
when v = 0, we are half way though the problem at the max height.
v = Vi - g t
0 = 7 - 9.8 t
t = .714 m/s at the top
so total time in air = 2 t = 1.43 seconds
how far did it go in that 1.43 seconds?
u = 14 cos 30 = 14 (sqrt 3) / 2 = 7 sqrt 3
so
d = range = 1.43 * 7 sqrt 3
d - 11 = distance that Gronk runs
so
speed times time = d - 11
s * 1.43 = d - 11
s = (d-11)/1.43
To find the constant velocity the receiver must have in order to catch the ball at the instant it is 1.6m above the ground, we can break this down into horizontal and vertical components.
First, let's find the time it takes for the ball to reach a height of 1.6m above the ground. We can use the vertical motion equation:
h = v₀y * t + (1/2) * g * t²
Where:
h = height above the ground (1.6m)
v₀y = initial vertical velocity (14 m/s * sin(30°))
t = time
g = acceleration due to gravity (9.8 m/s²)
Plugging in the values:
1.6m = (14 m/s * sin(30°)) * t + (1/2) * (9.8 m/s²) * t²
Simplifying the equation:
0 = 4.9t² + 6t - 1.6
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Plugging in the values:
t = (-6 ± √(6² - 4 * 4.9 * (-1.6))) / (2 * 4.9)
Simplifying further:
t ≈ 0.264s or t ≈ -1.827s
Since time cannot be negative, we discard the negative solution.
Now that we have the required time it takes for the ball to reach a height of 1.6m above the ground, we can use the horizontal motion equation to determine the constant velocity the receiver must have:
d = v₀x * t
Where:
d = distance (11m)
v₀x = initial horizontal velocity (14 m/s * cos(30°))
t = time (0.264s)
Plugging in the values:
11m = (14 m/s * cos(30°)) * 0.264s
Simplifying:
11m ≈ 3.83s
Therefore, the receiver must have a constant velocity of approximately 3.83 m/s along the line of flight of the ball to catch it at the instant it is 1.6m above the ground.
To solve this problem, we need to analyze the motion of both the football and the receiver separately. Let's break it down step by step:
1. First, let's analyze the motion of the football:
- We know that the initial vertical position of the football is 1.6m above the ground.
- The initial speed of the football is 14 m/s at an angle of 30 degrees above the horizontal.
- Ignoring any air resistance, we can break the initial velocity of the football into its horizontal and vertical components.
- The horizontal component (Vx) can be found using the formula Vx = V * cos(theta), where V is the initial speed and theta is the angle above the horizontal.
- The vertical component (Vy) can be found using the formula Vy = V * sin(theta).
- The horizontal motion of the football is uniform, so the horizontal velocity remains constant throughout the flight.
2. Next, let's analyze the motion of the receiver:
- The receiver starts 11m away from the quarterback along the line of flight of the ball.
- We want to find the constant velocity at which the receiver should move to catch the ball when it is 1.6m above the ground.
- Assuming that the receiver can move at a constant velocity throughout the motion, we need to calculate the required velocity.
Now, let's calculate the values needed to find the receiver's velocity:
1. Calculate the horizontal component of the football's initial velocity:
Vx = V * cos(theta) = 14 m/s * cos(30o) = 12.124 m/s
2. Calculate the vertical component of the football's initial velocity:
Vy = V * sin(theta) = 14 m/s * sin(30o) = 7 m/s
3. Find the time it takes for the football to reach the receiver's position:
- First, calculate the time it takes for the football to travel horizontally:
distance = velocity * time
11m = Vx * t
t = 11m / 12.124 m/s = 0.907 seconds (approximately)
4. Calculate the height of the football at that time:
- The equation for the vertical motion of the football is h = Vy * t + (1/2) * g * t^2, where g is the acceleration due to gravity (9.8 m/s^2).
- Plugging in the values:
h = 7 m/s * 0.907 s + (1/2) * 9.8 m/s^2 * (0.907 s)^2 = 5.47 m
5. Determine the time it takes for the receiver to reach the football's height (1.6m):
t_receiver = (1.6 m - 5.47 m) / (Vy) = (-3.87 m) / (7 m/s) = -0.553 seconds (approximately)
Since the obtained time is negative, it means that the receiver should already be at the required height (1.6m) when the football reaches that height. Therefore, there is no need to calculate the receiver's velocity.
So, to get to the ball at the instant it is 1.6m above the ground, the receiver must position himself at that height from the start.