a. Substitute the values into the equation:
0 = -16t^2 + 135t + 76
b. Using the quadratic formula, we have:
t = [-b ± √(b^2 - 4ac)] / 2a
Plugging in the values for a, b, and c from the equation:
t = [-135 ± √(135^2 - 4*(-16)*76)] / 2*(-16)
t = [-135 ± √(18225 + 4864)] / -32
t = [-135 ± √23089] / -32
t = [-135 ± 152.08] / -32
Therefore, the two values of t are:
t = (-135 + 152.08) / -32 ≈ 0.54 seconds
t = (-135 - 152.08) / -32 ≈ -16.68 seconds
Since time cannot be negative, the rocket will hit the ground approximately 0.54 seconds after being launched.
Use the quadratic formula to solve the equation. If necessary round to the nearest hundredth.
A rocket is launched from atop a 76-foot cliff with an initial velocity of 135 ft/s.
a. Substitute the values into the vertical motion formula h = -16t^2 + vt + c. Let h = 0.
b. Use the quadratic formula find out how long the rocket will take to hit the ground after it is launched. Round to the nearest tenth of a second.
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