A kicker punts a football from 3 feet above the ground with an initial velocity of 47 feet per second.
Write an equation that gives the height (in feet) of the football as a function of the time (in seconds) since it was punted.
Find the height (in feet) of the football 2 seconds after the punt.
Calculate how many seconds after the punt the ball would hit the ground.
height = -16t^2 + 47t + 3
when t=2
height = -64 + 94 + 3 = 33 feet high
to hit the ground, height = 0
-16t^2 + 47t+3=0
16t^2 - 47t - 3 = 0
(t-3)(16t + 1) = 0
t = 3 or t = a negative, which is not valid
it will hit the ground after 3 seconds
To write an equation that gives the height of the football as a function of time, we can use the equation for vertical motion under constant acceleration:
h(t) = h₀ + v₀t - (1/2)gt²
Where:
h(t) is the height at time t,
h₀ is the initial height (3 feet in this case),
v₀ is the initial velocity (47 feet per second in this case), and
g is the acceleration due to gravity (approximately 32.2 feet per second²).
1. Equation for the height of the football as a function of time:
h(t) = 3 + 47t - (1/2)32.2t²
2. To find the height of the football 2 seconds after the punt, substitute t = 2 into the equation:
h(2) = 3 + 47(2) - (1/2)32.2(2)²
h(2) = 3 + 94 - (1/2)32.2(4)
h(2) = 3 + 94 - 64.4
h(2) = 96.6 feet
The height of the football 2 seconds after the punt is 96.6 feet.
3. To find the time at which the ball hits the ground, we set h(t) = 0 and solve for t:
0 = 3 + 47t - (1/2)32.2t²
This is a quadratic equation in t. Using the quadratic formula or factoring, we find that the two possible solutions for t are approximately 4.442 seconds and -1.923 seconds. The negative value is not meaningful in this context, so we ignore it.
Therefore, the ball would hit the ground approximately 4.442 seconds after the punt.
To find the equation that gives the height of the football as a function of time, we can use the equation of motion for an object in free fall:
h(t) = h0 + v0t - (1/2)gt^2
Where:
h(t) is the height of the object at time t,
h0 is the initial height,
v0 is the initial velocity,
g is the acceleration due to gravity (approximately -32.2 ft/s^2 for objects near the Earth's surface),
t is the time elapsed.
In this case, the initial height (h0) is 3 feet, the initial velocity (v0) is 47 feet per second, and the acceleration due to gravity (g) is -32.2 feet per second squared. Therefore, the equation becomes:
h(t) = 3 + 47t - (1/2)(32.2)t^2
To find the height of the football 2 seconds after the punt, we can substitute t = 2 into the equation:
h(2) = 3 + 47(2) - (1/2)(32.2)(2)^2
= 3 + 94 - (1/2)(32.2)(4)
= 3 + 94 - 64.4
= 97.6 feet
Therefore, the height of the football 2 seconds after the punt is 97.6 feet.
To calculate how many seconds after the punt the ball would hit the ground, we need to find the time (t) when the height (h(t)) is equal to 0, as the ball hits the ground at h(t) = 0. We can rewrite the equation with h(t) = 0 and solve for t:
0 = 3 + 47t - (1/2)(32.2)t^2
This is a quadratic equation, which can be solved using various methods. In this case, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
Where a, b, and c are the coefficients of the quadratic equation. In this case, a = -16.1, b = 47, and c = 3.
t = (-47 ± sqrt(47^2 - 4(-16.1)(3))) / 2(-16.1)
= (-47 ± sqrt(2209 + 193.2)) / -32.2
= (-47 ± sqrt(2402.2)) / -32.2
We only consider the positive value of t, as time cannot be negative in this context.
t ≈ 4.98 seconds
Therefore, the ball would hit the ground approximately 4.98 seconds after the punt.