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 101-foot 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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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 101-foot 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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after a substitution, we get: \[\Large h = - 16{t^2} + 116t + 101\]
your answer is correct.
the rocket will reach its maximum height at time: \[\Large {t_1} = \frac{{{v_0}}}{g} = \frac{{116}}{{32}} = 3.625\sec \]

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and it will return on the earth surface at time: \[\large {t_2} = \sqrt {{{\left( {\frac{{{v_0}}}{g}} \right)}^2} + \frac{{2{h_0}}}{g}} = 4.411\sec \]
so total time is: \[\Large 3.625 + 4.411 \cong 8.04\sec \]
so, you are right!
I have applied the equation of dynamic!
thank you both :)
:)
@Michele_Laino you are good in Physics. ^__^
thanks! :) @ASAAD123

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