An object is thrown from the top of an 80 foot building with an initial velocity of 64 feet per second. The height (h) of the object after (t) seconds is given by the quadratic equation h=-16t+64t+80 .When will the object hit the ground?
The 16t should be 16t^2. Check your Eq for errors.
d = 64t +16t^2 = 80
16t^2 + 64t - 80 = 0
(T-1)(T+5) = 0.
T = 1, and -s.
T = 1 s.
Correction:
16t^2 + 64t - 80 = 0
Divide both side by 16:
t^2 + 4t - 5 = 0.
(T-1)(t+5) = 0
T = 1, and -5.
Use positive value of T.
6.4
To find when the object will hit the ground, we need to determine the time at which the height (h) is equal to zero. We can do this by setting the equation h = 0 and solving for t.
The given equation is: h = -16t^2 + 64t + 80
Setting h = 0, we get: 0 = -16t^2 + 64t + 80
To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 64, and c = 80.
Substituting these values into the quadratic formula:
t = ( -64 ± √(64^2 - 4(-16)(80)) ) / (2(-16))
Simplifying further:
t = ( -64 ± √(4096 + 5120) ) / (-32)
t = ( -64 ± √9216 ) / (-32)
t = ( -64 ± 96 ) / (-32)
Now we have two possible solutions for t:
1. t = ( -64 + 96 ) / (-32) = 32 / (-32) = -1
2. t = ( -64 - 96 ) / (-32) = -160 / (-32) = 5
Since time cannot be negative in this context, we discard the negative solution t = -1.
Therefore, the object will hit the ground after 5 seconds.