The football is punted with an initial velocity of 22 meters per second from a height of 1 meter off the ground. When is the football within 5 metes of the ground?
Is the ball punted straight up? If so, then
h = 1+22t-4.9t^2
So, just solve for t when h<=5.
If the ball was kicked at an angle, things get a bit harder, as the equation is then
h = 1 + 22sinθ t - 4.9t^2
To find out when the football is within 5 meters of the ground, we can use the equation of motion for vertical displacement:
h = h0 + v0 * t + (1/2) * g * t^2
Where:
- h is the height of the object at time t
- h0 is the initial height of the object
- v0 is the initial velocity of the object
- g is the acceleration due to gravity (approximately -9.8 m/s^2)
- t is the time elapsed
In this case, we want to find the time when the football is within 5 meters of the ground, so we can set h (height) to 5 meters and solve for t.
5 = 1 + 22 * t + (1/2) * (-9.8) * t^2
Now, let's rearrange the equation and solve for t:
0 = -4.9 * t^2 + 22 * t - 4
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
a = -4.9, b = 22, c = -4
Substituting the values into the formula:
t = (-(22) ± √((22)^2 - 4*(-4.9)*(-4))) / (2*(-4.9))
Simplifying the equation further, we get:
t = (-22 ± √(484 - 78.4)) / (-9.8)
t = (-22 ± √(405.6)) / (-9.8)
t ≈ (-22 ± 20.138) / (-9.8)
We have two possible solutions:
t1 ≈ (-22 + 20.138) / (-9.8) ≈ 0.186 seconds
t2 ≈ (-22 - 20.138) / (-9.8) ≈ 4.43 seconds
Therefore, the football is within 5 meters of the ground approximately 0.186 seconds and 4.43 seconds after it is punted.