Question

A rocket is launched vertically from the ground with an initial velocity of 56 ft/sec

Answer 1

Answer:

y = 17x - 4.5x²

height = 16m at 2 sec

total time = 3.77 sec

Step-by-step explanation:

Given the initial velocity of rocket  = 56ft/sec

We use newton formula

s = ut + 1/2at²

where s = distance

             u = initial velocity

             t= time

             a = acceleration due to gravity = 9.8m/s²

since the rocket is launched upward, it is against the gravity so a = -9.81.

56 feet  = 17.0688 = 17

suppose s = height = y and time = x

y = 17x - 1/2(9.81)x²

or

y = 17x - 4.5x²

b)

as x can not be negative so domain will be from 0 to 3.78

put value of x to find value of y and then plot these on graph

This graphs will be a parabola which opens downward.

c)

at 2 sec, y value is greatest = 16 meters

time for which rocket was in the air = 3.77sec

Answer 2

Answer:

a) h(t) = -16t² + 56t; b) the graph is attached; c) maximum height is 49 feet, it takes 1.75 seconds to reach the maximum, and it is in the air for 3.5 seconds.

Step-by-step explanation:

Equations for these situations are in the form

h(t) = -16t² + v₀t + h₀, where t is the tie in seconds, v₀ is the initial velocity and h₀ is the initial height. The initial height in this case is 0, since it is launched from the ground; and the initial velocity is 56.  This gives us the function

h(t) = -16t² + 56t

The graph is attached.

Using the graph, we can trace the function to the maximum.  The y-value of the maximum, or the maximum height, is 49 feet.  The x-value of the maximum, the time it takes, is 1.75 seconds.

The roots of this equation are at x = 0 and x = 3.5.  This means the rocket is in the air for 3.5-0 = 3.5 seconds.