Answer:
The time that the rocket will hit the ground is after 2.78 seconds.
Step-by-step explanation:
You know that the height (H) is given by the formula H= −16*t² + v*t+ h
A rocket is launched directly from the side of a 40 m high cliff and that the initial velocity is 30 m / sec. So this indicates that the value of v is 30 m / sec and h is the height of 40 m. Replacing this value in the formula you get:
H= −16*t² + 30*t+ 40
You must find the moment when the rocket will hit the ground, that is, the moment when the height H is zero:
0= −16*t² + 30*t+ 40
To calculate the time t the expression is used:
[tex]\frac{-b+-\sqrt{b^{2}-4*a*c } }{2*a}[/tex]
Being the general quadratic equation ax² + bx + c = 0, in this case a=-16, b=30 and c= 40. Replacing:
[tex]\frac{-30+-\sqrt{30^{2}-4*(-16)*40} }{2*(-16)}[/tex]
Solving:
[tex]\frac{-30+-\sqrt{900+2,560} }{2*(-16)}[/tex]
[tex]\frac{-30+-\sqrt{3,460} }{2*(-16)}[/tex]
[tex]\frac{-30+-58.82 }{2*(-16)}[/tex]
Then:
[tex]t1=\frac{-30+58.82 }{2*(-16)}[/tex] or [tex]t2=\frac{-30-58.82 }{2*(-16)}[/tex]
Solving for t1:
[tex]t1=\frac{28.82 }{2*(-16)}[/tex]
[tex]t1=\frac{28.82 }{-32}[/tex]
t1=-0.9
Solving for t2:
[tex]t2=\frac{-88.82 }{2*(-16)}[/tex]
[tex]t2=\frac{-88.82 }{-32}[/tex]
t2=2.78
Since the time cannot be negative, the time that the rocket will hit the ground is after 2.78 seconds.
Time cannot be negative therefore, the correct answer is (t = 2.78) and this can be determined by using the given data.
Given :
When a rocket hit the ground that means, H = 0 and the equation becomes:
[tex]-16t^2+30t+40=0[/tex]
[tex]t = \dfrac{-30\pm\sqrt{(30)^2-(4\times -16\times 40)} }{2\times -16}[/tex]
[tex]t=\dfrac{-30\pm58.82}{-32}[/tex]
t = -0.9, 2.78
Time cannot be negative therefore, the correct answer is (t = 2.78).
For more information, refer to the link given below:
https://brainly.com/question/21835898
