Question

NASA launches a rocket at

Answer 1

Answer:

t = 28.3 seconds

1200 meters

Step-by-step explanation:

The height of the rocket after t seconds from launching is given by

h(t) = - 4.9t² + 124t +416

Now, at h = 0 the rocket will splashes down.

Hence, -4.9t² + 124t + 416 = 0

Applying Sridhar Acharya formula,

t = $\frac{-124 -\sqrt{124^{2}- 4 \times(-4.9) \times416 } }{2 \times (-4.9)}$ or, t = $\frac{-124 +\sqrt{124^{2}- 4 \times(-4.9) \times416 } }{2 \times (-4.9)}$

Hence, t = 28.3 or, t = -2.999

Since t can not be negative, so, t = 28.3 seconds. (Answer)

Now, h = -4.9t² + 124t + 416, for h to be maximum $\frac{dh}{dt} =0$

⇒ -9.8t + 124 = 0

⇒ t =12.65 secs

Hence, $h(x)_{max} =-4.9 \times(12.65)^{2} +124 \times(12.65)+416 =1200$ meters

Therefore, the rocket peaks at 1200 meters above sea level. (Answer)