Rachael shoots a rocket straight upward with an initial velocity of 4 meters per second from a height of 180 meters. The distance [tex]\( d \)[/tex] (in meters) of the rocket from the ground after [tex]\( t \)[/tex] seconds is given by the equation

[tex]\[ d = 180 + 4t - 0.5g \left( t^2 \right) \][/tex]

where gravity [tex]\( g \)[/tex] is [tex]\( 9.8 \, m/s^2 \)[/tex]. After approximately how many seconds does the rocket strike the ground?

A. 5 seconds
B. 6.5 seconds
C. 8.5 seconds
D. 9 seconds
E. 10 seconds



Answer :

Sure, let's solve this problem step-by-step.

First, we start with the given equation for the distance [tex]\(d\)[/tex] from the ground:
[tex]\[ d = 180 + 4t - 0.5 \cdot 9.8 \cdot t^2 \][/tex]

Simplifying the equation, we get:
[tex]\[ d = 180 + 4t - 4.9t^2 \][/tex]

We need to find the time [tex]\(t\)[/tex] when the rocket strikes the ground. When the rocket is on the ground, [tex]\(d = 0\)[/tex]. Therefore, we set the equation to 0:
[tex]\[ 0 = 180 + 4t - 4.9t^2 \][/tex]

Rearranging the terms to get it in the standard quadratic equation form [tex]\(at^2 + bt + c = 0\)[/tex]:
[tex]\[ -4.9t^2 + 4t + 180 = 0 \][/tex]

Here, [tex]\(a = -4.9\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = 180\)[/tex].

To solve this quadratic equation, we use the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ t = \frac{-4 \pm \sqrt{4^2 - 4 \cdot (-4.9) \cdot 180}}{2 \cdot (-4.9)} \][/tex]

First, calculate the discriminant (the part under the square root):
[tex]\[ \Delta = 4^2 - 4 \cdot (-4.9) \cdot 180 \][/tex]
[tex]\[ \Delta = 16 + 4 \cdot 4.9 \cdot 180 \][/tex]
[tex]\[ \Delta = 16 + 3528 \][/tex]
[tex]\[ \Delta = 3544 \][/tex]

Next, solve for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{-4 \pm \sqrt{3544}}{-9.8} \][/tex]
[tex]\[ \sqrt{3544} \approx 59.56 \][/tex]

Thus, we have two solutions for [tex]\(t\)[/tex]:
[tex]\[ t_1 = \frac{-4 + 59.56}{-9.8} \][/tex]
[tex]\[ t_1 \approx \frac{55.56}{-9.8} \][/tex]
[tex]\[ t_1 \approx -5.67 \, \text{seconds} \][/tex]

[tex]\[ t_2 = \frac{-4 - 59.56}{-9.8} \][/tex]
[tex]\[ t_2 \approx \frac{-63.56}{-9.8} \][/tex]
[tex]\[ t_2 \approx 6.48 \, \text{seconds} \][/tex]

Since time cannot be negative, we discard [tex]\(t_1\approx -5.67\)[/tex] and consider [tex]\(t_2 \approx 6.48\)[/tex].

We'll now round [tex]\(6.48\)[/tex] to the nearest given option. The options are: 5, 6.5, 8.5, 9, 10.

The closest option to [tex]\(6.48\)[/tex] is [tex]\(6.5\)[/tex].

Therefore, the rocket strikes the ground after approximately:
(B) 6.5 seconds