Different initial up speed but use :
http://www.jiskha.com/display.cgi?id=1396965991#1396965991.1396967805
s = v0t − 16t2,
where s is the height, v0 is the initial velocity, and t is the time. Find the time for a projectile to return to Earth if it has an initial velocity of 184 ft/s.
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http://www.jiskha.com/display.cgi?id=1396965991#1396965991.1396967805
The formula for the height of the projectile is given as:
s = v0t - 16t^2
If the projectile returns to Earth, its height will be zero, so we can set s = 0 and solve for t.
0 = v0t - 16t^2
Rearranging the equation, we have:
16t^2 = v0t
Divide both sides of the equation by t:
16t = v0
Now, substitute the given value for the initial velocity, v0 = 184 ft/s:
16t = 184
Divide both sides of the equation by 16:
t = 184/16
Simplifying the right side of the equation:
t = 11.5
Therefore, the time for the projectile to return to Earth is 11.5 seconds.
In this case, the projectile is returning to Earth, so the height will be 0. Therefore, we can rewrite the equation as:
0 = v0t - 16t^2
Let's plug in the given initial velocity, v0 = 184 ft/s:
0 = 184t - 16t^2
This is a quadratic equation in terms of t. We can now use the quadratic formula to solve for t. The quadratic formula states:
t = (-b ± √(b^2 - 4ac)) / (2a)
Comparing our equation with the standard quadratic form (ax^2 + bx + c = 0), we have:
a = -16, b = 184, and c = 0.
Substituting these values into the quadratic formula, we get:
t = (-184 ± √(184^2 - 4*(-16)*0)) / (2*(-16))
Simplifying further:
t = (-184 ± √(33856)) / (-32)
Finally, solving for t gives us two possible solutions:
t = (-184 + √(33856)) / (-32)
t = (-184 - √(33856)) / (-32)
Calculating these values will give us the times for the projectile to return to Earth.