A baseball is thrown up in the air from a height of 3 feet with an initial velocity of 23 feet per second. When does the baseball hit the ground?
hf=hi+vi*t-16t^2
0=3+23*t-16t^2
this is a quadratic, solve it with the quadratic equation. I find it odd that Old English units (feet) are being used here, you must have a very old text, or a very old teacher.
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To determine when the baseball hits the ground, we need to use the equations of motion to calculate the time it takes for the baseball to reach the ground. The equation we will use is:
y = h + v0t - 1/2gt^2
Where:
- y is the final height of the baseball (which is 0 when it hits the ground)
- h is the initial height of the baseball (3 feet in this case)
- v0 is the initial velocity of the baseball (23 feet per second)
- g is the acceleration due to gravity (32.2 feet per second squared)
- t is the time it takes for the baseball to hit the ground
Using the equation of motion, we can rewrite it as:
0 = 3 + 23t - 1/2(32.2)t^2
Now, we need to solve this quadratic equation for t. Rearranging the equation, we get:
16.1t^2 - 23t + 3 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 16.1, b = -23, and c = 3. By substituting these values into the quadratic formula, we get:
t = (-(-23) ± √((-23)^2 - 4(16.1)(3))) / (2(16.1))
Simplifying further, we have:
t = (23 ± √(529 - 193.2)) / 32.2
t = (23 ± √335.8) / 32.2
Now, we can calculate the two possible values for t:
t1 = (23 + √335.8) / 32.2 ≈ 2.9 seconds
t2 = (23 - √335.8) / 32.2 ≈ 0.1 seconds
Since time cannot be negative, we discard t2 as it is not a valid solution. Therefore, the baseball hits the ground approximately 2.9 seconds after being thrown.