A football is thrown down the field by a quarterback from a height of 1 foot with an initial velocity of 40 feet per second at angle of pi/6 radians. If a receiver catches the ball downfield at a height of 1 foot above ground;
a) How long was the ball in the air?
b) What was the maximum height of the ball?
recall that the height
y = 1 + 40*sin(pi/6) t - 8t^2
(a) solve for t when y=1
(b) as with all parabolas, the vertex is at t = -b/2a
the given height equation is incorrect
... the t^2 coefficient is -16, not -8
solve away...
correct. good catch
To answer these questions, we first need to break down the given information and use some basic principles of projectile motion.
a) How long was the ball in the air?
In order to find the time the ball is in the air, we can use the vertical motion of the ball. We know that the initial vertical velocity is given by V0y = V0 * sin(theta), where V0 is the initial velocity (40 ft/s) and theta is the angle of projection (pi/6 radians).
Now, we can use the equation of motion for vertical displacement:
y = V0y * t - (1/2) * g * t^2
Where y is the vertical displacement, t is the time, and g is the acceleration due to gravity (approximately 32.2 ft/s^2). The initial vertical displacement y is 1 ft, and we want to find the time t when the vertical displacement is also 1 ft.
Plugging in the values:
1 = V0y * t - (1/2) * g * t^2
Substituting the values of V0y and g, we get:
1 = (40 * sin(pi/6)) * t - (1/2) * 32.2 * t^2
Simplifying further:
1 = 20 * t - 16.1 * t^2
Rearranging the equation to get a quadratic equation in the standard form:
16.1 * t^2 - 20 * t + 1 = 0
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values:
t = (-(-20) ± √((-20)^2 - 4 * 16.1 * 1)) / (2 * 16.1)
t = (20 ± √(400 - 64.4)) / 32.2
Simplifying further:
t ≈ 2.92 seconds or t ≈ 0.05 seconds
Since it does not make sense for the ball to have a negative time, we can ignore the negative root. Therefore, the ball was in the air for approximately 2.92 seconds.
b) What was the maximum height of the ball?
To find the maximum height of the ball, we need to determine the time it takes for the ball to reach its highest point. At the highest point, the vertical velocity becomes zero. We can use the equation for the vertical component of velocity:
V = V0y - g * t
Setting V equal to zero and solving for t:
0 = (40 * sin(pi/6)) - 32.2 * t
Simplifying further:
t = (40 * sin(pi/6)) / 32.2
t ≈ 0.752 seconds
Now, we can use this time to find the maximum height by substituting it back into the equation for vertical displacement:
y = V0y * t - (1/2) * g * t^2
y = (40 * sin(pi/6)) * 0.752 - (1/2) * 32.2 * (0.752)^2
Simplifying further, we can calculate the maximum height:
y ≈ 10.08 feet
Therefore, the maximum height of the ball is approximately 10.08 feet.