A rocket is launched from atop a 105ft cliff with initial velocity of 156 ft/s Height above ground a time t is given -16t^2 + 156t + 105 = h. When will it hit the ground after launch?
clearly, when
-16t^2 + 156t + 105 = 0
To find out when the rocket will hit the ground after launch, we need to determine the time at which its height above the ground becomes zero.
Given the equation for the height of the rocket above the ground at time t: -16t^2 + 156t + 105 = h, we can substitute h with 0 to find the time when the rocket hits the ground.
So, we have the equation: -16t^2 + 156t + 105 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -16, b = 156, and c = 105.
Plugging in the values, we get:
t = (-(156) ± √((156)^2 - 4(-16)(105))) / (2(-16))
Simplifying further:
t = (-156 ± √(24336 + 6720)) / (-32)
t = (-156 ± √31056) / (-32)
Now, let's calculate the values under the square root:
√31056 ≈ 176.034
Now, substitute this value into the quadratic formula to find the two possible values of t:
t1 = (-156 + 176.034) / (-32)
t2 = (-156 - 176.034) / (-32)
Simplifying further:
t1 ≈ 0.438
t2 ≈ 9.062
Therefore, the rocket will hit the ground approximately 0.438 seconds and 9.062 seconds after launch.