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


Comments

Popular posts from this blog

Papaano Natamasa Ng Japan Ang Moderninasyon?

Paano Na Aapektuhan Ang Mga Gawain Nang Mnlf Ang Ekonomiya

Konklusyon At Rekomendasyon Sa Pagtaas Ng Presyo Sa Pamilihan