Use The Quadratic Formula To Solve The Equation. If Necessary, Round To The Nearest Hundredth., A Rocket Is Launched From Atop A 55-Foot Cliff With An

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

A rocket is launched from atop a 55-foot cliff with an initial velocity of 138 ft/s.
a. Substitute the values into the vertical motion formula mc012-1.jpg. 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.

(Hello!  I can only solve its time to hit the ground.)

Answer:

The time rocket took to hit the ground after it is launched is approximately at 8.16 seconds.

Step-by-step explanation:

Given:

y = 55 ft = 16.764 m

V_{y0} = 138 ft = 42.0624 m

a_{y} = -9.81 \frac{m}{s^{2}}

Formula:

Using this constant-acceleration equation y = V_{y0} t + \frac{1}{2} a_{y} t^{2} and the quadratic formula \frac{-b+\sqrt{b^{2}-4ac } }{2a} ; \frac{-b-\sqrt{b^{2}-4ac } }{2a}, by getting the highest positive answer only, we can find the time it took.

Equation:

y = V_{y0} t + \frac{1}{2} a_{y} t^{2}

\frac{-b+\sqrt{b^{2}-4ac } }{2a} ; \frac{-b-\sqrt{b^{2}-4ac } }{2a}

Solution:

y = V_{y0} t + \frac{1}{2} a_{y} t^{2}

16.764 m = 42.0624 m t + \frac{1}{2} -9.81 \frac{m}{s^{2}} t^{2}

16.764 m = 42.0624 m t - 4.905 \frac{m}{s^{2}} t^{2}

0 = -16.764 m + 42.0624 m t - 4.905 \frac{m}{s^{2}} t^{2}

(0 = -16.764 m + 42.0624 m t - 4.905 \frac{m}{s^{2}} t^{2}) (-1)

0 = 16.764 m - 42.0624 m t + 4.905 \frac{m}{s^{2}} t^{2}

where:

a = 4.905 \frac{m}{s^{2}} t^{2} = 4.905

b = -42.0624 m t = - 42.0624

c = 16.764 m = 16.764

t = 8.156386993 s

Answer:

t = 8.156386993 s


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