Please explain to work this:
A rocket is launched from atop a 105-foot cliff with an initial velocity of 156ft/s. the height of the rocket above the ground at time t is given by
h= -16t^2 + 156t + 105. When will the
rocket hit the ground after it is launched? Round to the nearest second.
naturally, when h = 0.
-16t^2 + 156t + 105 = 0
t = (39 ± √1941)/8
t = 10 sec
???????????????????????????????????????Your confusing me! STOP CONFUSING ME!!!!!
h=0, so you set this up in the quadratic formula, which then you follow all the rules and your left with a positive and negative answer. time cant be negative, so the correct answer should be about 10.38 seconds which rounds off to 10.4
I agree with anonymous
To find when the rocket hits the ground, we need to determine the value of t when the height (h) is equal to 0.
In the given equation, the height of the rocket above the ground (h) is given by:
h = -16t^2 + 156t + 105
Setting h to zero and solving for t:
0 = -16t^2 + 156t + 105
This equation is a quadratic equation, so we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 156, and c = 105.
Now we can plug these values into the quadratic formula:
t = (-156 ± √(156^2 - 4*(-16)*105)) / (2*(-16))
Simplifying further:
t = (-156 ± √(24336 + 6720)) / (-32)
t = (-156 ± √(31056)) / (-32)
t = (-156 ± 176) / (-32)
Now we can calculate the two possible values for t:
t1 = (-156 + 176) / (-32) = 20 / (-32) = -0.625
t2 = (-156 - 176) / (-32) = -332 / (-32) = 10.375
Since time cannot be negative in this context, we can ignore the negative value of t (-0.625).
Therefore, the rocket will hit the ground approximately 10 seconds after it is launched.