If an object is launched straight up into the air from a starting height of h_{0} feet, then the height of the object after t seconds is approximately h=-16t^2+v_{0}t+h_{0} feet, where v_{0} is the initial velocity of the object. Find the starting height and initial velocity of an object that attains a maximum height of 412 feet five seconds after being launched.
To find the starting height and initial velocity of the object, we can use the given information that the object attains a maximum height of 412 feet five seconds after being launched.
We know that the height of the object at any time t is given by the equation h = -16t^2 + v0t + h0.
At the maximum height, the object's velocity is zero. Therefore, we can determine the time when the object reaches its maximum height by finding the time t when the velocity term (-16t^2 + v0t) becomes zero.
Since the object attains its maximum height five seconds after being launched, we can substitute t = 5 into the equation and solve for v0.
0 = -16(5^2) + v0(5) + h0
0 = -400 + 5v0 + h0
Now, we're given that the maximum height is 412 feet, so we can set h0 = 412 and solve for v0.
0 = -400 + 5v0 + 412
-12 = 5v0
v0 = -2.4
Therefore, the initial velocity of the object is -2.4 feet/second.
To find the starting height, we substitute the values of v0 and t into the equation and solve for h0.
h = -16(5^2) + (-2.4)(5) + h0
412 = -400 - 12 + h0
412 = -412 + h0
h0 = 824
Therefore, the starting height of the object is 824 feet.