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.
1.)0= -16t^2 + 135t + 76; 0.5 s
2.)0= -16t^2 + 135t + 76; 9 s

A ) h = -16 t² + 135 t + 76
Let : h = 0
0 = - 16 t² + 135 t + 76
B ) t 1/2 = (-b+/- √ ( b² - 4 ac ) / ( 2 a )
t 1/2 = (-135 - √(18,225 + 4,864))/ (-32) = ( - 135 - 151.95) / (- 32)=
= (-286.95) / (- 32) = 9.967 ≈ 9.0 s ( other solution is negative )
Answer:
2) 0 = -16 t² + 135 t + 76; 9 s

Answer Link

wagonbelleville wagonbelleville · Answer 2 · 2019-01-03T13:04:10+01:00

Answer:

The answer is 2.) [tex]0=-16t^{2}+135t+76;9 s[/tex]

Step-by-step explanation:

Given initial velocity=135 ft/s

& cliff=76 foot

Given quadratic equation

[tex]0=-16t^{2}+vt+c[/tex]

⇒ [tex]0=-16t^{2}+135t+76[/tex] (let h=0 it is given)

⇒ [tex]t=\frac{-135\pm\sqrt ((135)^{2}-4(-16)(76))}{2(-16)}[/tex]

⇒ [tex]t=\frac{-135\pm151.951}{-32}[/tex]

⇒ t=8.96≈9 s (the other root is negative)

Hence, rocket will take 9 s to hit the ground after launched.