an object is thrown upward from the top of a 80-foot building with a initial velocity of 65 feet per second. The height h of the object after t seconds is given by the equation h=-16t2+64t+80. when will the object hit the ground
Your equation does not match your data
if the equation is
h = -16t^2 + 64t + 80 , then the initial velocity was 64 ft/sec, not 65 like you typed.
I will go with the 64
-16t^2 + 64t + 80 = 0
divide each term by -16
t^2 - 4t - 5 = 0
(t+1)(t-5) = 0
t = 5 or -1 , we will reject the last answer
the object will hit the ground after 5 seconds
To find the time at which the object hits the ground, we need to solve the equation h = -16t^2 + 64t + 80 for when h = 0.
So, we have the equation:
-16t^2 + 64t + 80 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 64, and c = 80. Let's substitute these values into the quadratic formula and solve for t:
t = (-64 ± √(64^2 - 4(-16)(80))) / (2(-16))
t = (-64 ± √(4096 + 5120)) / (-32)
t = (-64 ± √(9216)) / (-32)
t = (-64 ± 96) / (-32)
Simplifying further, we have two possible solutions:
t1 = (-64 + 96) / (-32) = 1 second
t2 = (-64 - 96) / (-32) = 5 seconds
Therefore, the object will hit the ground twice, once at 1 second and again at 5 seconds.
To find when the object hits the ground, we need to determine the value of t when the height h is equal to zero.
The given equation for the height of the object is h = -16t^2 + 64t + 80.
Setting h to zero, we can solve the quadratic equation -16t^2 + 64t + 80 = 0.
To simplify the equation, we can divide through by -8 to make the coefficient of the t^2 term positive: 2t^2 - 8t - 10 = 0.
Now we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. In this case, factoring will not work, so let's use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a.
For our equation 2t^2 - 8t - 10 = 0, the coefficients are a = 2, b = -8, and c = -10.
Substituting these values into the quadratic formula, we get: t = (-(-8) ± √((-8)^2 - 4 * 2 * -10)) / (2 * 2).
Simplifying further, we have: t = (8 ± √(64 + 80)) / 4.
Now, we calculate the value within the square root: √(64 + 80) = √144 = 12.
Substituting this back into the equation, we have: t = (8 ± 12) / 4.
Now, we can consider both possibilities:
1. t = (8 + 12) / 4 = 20 / 4 = 5. This is one possible solution.
2. t = (8 - 12) / 4 = -4 / 4 = -1. This is the other possible solution.
Since time cannot be negative in this context, we discard the negative solution, which means t = 5.
Therefore, the object will hit the ground after 5 seconds.