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anonymous
 one year ago
Check on my answer please
Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.
A rocket is launched from atop a 101foot cliff with an initial velocity of 116 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.
A) 0 = 16t^2+101t+116; 8s
B) 0 = 16t^2+116t+101; 0.8s
C) 0 = 16t^2+101t+116; 0.8s
D) 0 = 16t^2+116t+101; 8s  my answer
anonymous
 one year ago
Check on my answer please Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth. A rocket is launched from atop a 101foot cliff with an initial velocity of 116 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. A) 0 = 16t^2+101t+116; 8s B) 0 = 16t^2+116t+101; 0.8s C) 0 = 16t^2+101t+116; 0.8s D) 0 = 16t^2+116t+101; 8s  my answer

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1after a substitution, we get: \[\Large h =  16{t^2} + 116t + 101\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your answer is correct.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the rocket will reach its maximum height at time: \[\Large {t_1} = \frac{{{v_0}}}{g} = \frac{{116}}{{32}} = 3.625\sec \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and it will return on the earth surface at time: \[\large {t_2} = \sqrt {{{\left( {\frac{{{v_0}}}{g}} \right)}^2} + \frac{{2{h_0}}}{g}} = 4.411\sec \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so total time is: \[\Large 3.625 + 4.411 \cong 8.04\sec \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so, you are right!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I have applied the equation of dynamic!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino you are good in Physics. ^__^

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1thanks! :) @ASAAD123
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