A kicker punts a football from 3 feet above the ground with an initial velocity of 47 feet per second.
1. Write an equation that gives the height (in feet) of the football as a function of the time (in secs) since it was punted.
Okay. According to my text book, h = -16t^2 + vt + s (T= secs, V= vertical velocity, S= initial height)
So, I entered the info into the equation and got
h =-16t^2 + 47t +3
That's what they're looking for, right?
2. Find the height (in feet) of the football 2 seconds after the punt.
So, I entered 2 in the equation and got
h = -16(2)^2 + 47(2) + 3
h = -32^2 + 97 = -927ft
But how can the height be negative?
I must have done something wrong.
3. Calculate how many seconds after the punt the ball would hit the ground.
Now I'm lost.
Any help you can provide would be great. Thanks!
Ahem.
h = -16(2)^2 + 47(2) + 3
You do not have -(16*2)^2, you just have -16 * 2^2 = -16*4
So,
h(2) = -64 + 97 = 33
Ball hits the ground when h=0, so
-16t^2 + 47t + 3 = 0
-(t-3)(16t+1) = 0
t = 3
Thanks.
1. Yes, you are correct in using the equation h = -16t^2 + vt + s, where h is the height, t is the time in seconds, v is the vertical velocity (or initial velocity), and s is the initial height.
2. To find the height of the football 2 seconds after the punt, you can substitute t = 2 into the equation:
h = -16(2)^2 + 47(2) + 3
h = -64 + 94 + 3
h = 33 feet
So, the height of the football 2 seconds after the punt is 33 feet. It's important to note that the height can be negative if the ball has already hit the ground.
3. To calculate how many seconds after the punt the ball would hit the ground, you need to find the time when h = 0. Set the equation equal to 0 and solve for t:
-16t^2 + 47t + 3 = 0
There are different methods to solve this quadratic equation, but one common way is to use the quadratic formula: t = (-b ± √(b^2 - 4ac))/(2a), where a, b, and c are the coefficients of the quadratic equation. In this case, a = -16, b = 47, and c = 3.
Applying the quadratic formula:
t = (-47 ± √(47^2 - 4(-16)(3)))/(2(-16))
t = (-47 ± √(2209 + 192))/(2(-16))
t = (-47 ± √2401)/(2(-16))
Since we are considering the time after the punt, it's safe to use the positive answer from the ± operation:
t = (-47 + √2401)/(2(-16))
t = (-47 + 49)/(2(-16))
t = 2/(-32)
t = -1/16
The negative value for t is because we're measuring time relative to the moment of the punt. However, since time cannot be negative in this context, we can conclude that the ball does not hit the ground within the time frame of the equation.
Therefore, the ball does not hit the ground within the time frame given by the equation.