Question

Tyrone launches a toy rocket into the air with an initial upward velocity of 49 ft/s and an initial height of 0 ft. a. How long will it take the rocket to reach its maximum height? Round to the nearest hundredth. b. How high above the ground will it be? Round to the nearest tenth. c. What is the range of the function? d. Why is there a maximum point on the graph of this function?

Answer 1

a) 1.52 s

b) 37.3 ft

c) $[0,37.3] ft$

d) Because it has a maximum whose value is 37.3 ft

Step-by-step explanation:

a)

The motion of the toy rocket is a free fall motion (it is acted upon the force of gravity only), so it is subjected to a constant acceleration of

$g=32.2 ft/s^2$

towards the ground.

Therefore, the velocity of the rocket after time t is given by the following suvat equation:

$v=u-gt$

where

v is the final velocity

u is the initial velocity

t is the time

The rocket reaches the maximum height when the velocity becomes zero, so when

v = 0

Also, the initial velocity is

u = 49 ft/s

Therefore, solving the equation for t, we find the time:

$0=u-gt\\t=\frac{u}{g}=\frac{49}{32.2}=1.52 s$

b)

In a motion with constant acceleration in free fall, the displacement of the object is given by

$s=ut-\frac{1}{2}gt^2$

where

u is the initial velocity

t is the time

g is the acceleration due to gravity

s is the displacement

In this problem, we have:

$u=49 ft/s$ is the initial vertical velocity

t = 1.52 is the time taken to reach the maximum height

$g=32.2 ft/s^2$ is the acceleration due to gravity

So, the maximum height reaches by the rocket is:

$s=(49)(1.52)-\frac{1}{2}(32.2)(1.52)^2=37.3 ft$

c)

The function that describes the height of the rocket at time t is:

$h(t) = h_0 + ut - \frac{1}{2}gt^2$

where

$h_0 = 0ft$ is the initial height

t is the time

u = 49.7 ft/s is the initial velocity

$g=32.2 ft/s^2$ is the acceleration due to gravity

Substituting, we find the explicit expression:

$h(t) = 49t-16.1t^2$

The range of the a function corresponds to the set of values that the function (in this case, h(t)) can take.

In this problem, we know that the maximum value of h(t) is

$h_{max}=37.3 ft$ (part c)

Also, the value of h(t) must be positive, since the height of the rocket cannot be negative, so the minimum value is zero:

$h_{min}=0$

So the range of the function is:

$[0,37.3] ft$

d)

This function represented on a graph would show a downward parabola, with maximum value at 37.3 ft.

The value of the maximum represents the maximum height of the rocket, as calculated in part b: therefore, the reason for which there is a maximum point on the graph of the function is that the function h(t) has a maximum at 37.3 ft, so it cannot assume larger values.

Answer 2

a) The time is taken to reach maximum height is 1.52 sec.

b) The highest point is 37.28 ft.

c)  The range of the function is from 0 to 37.28 ft.

d) A maximum point on the graph of this function is 37.28 ft.

What is projectile motion?

When an object is thrown at an angle then it traces the path of the parabola is called projectile motion.

Given

Tyrone launches a toy rocket into the air with an initial upward velocity (u) of 49 ft/s and an initial height of 0 ft.

Then we know that the equation of motion

$\rm v= u + at \ \ \ ...1\\s = ut + \dfrac{1}{2} a t ^2 \ \ \ ...2\\v^2 = u^2 + 2as \ \ \ ...3$

where

v = final velovity

u = initial velocity

s = displacement

a = acceleration

a) The time is taken to reach maximum height.

At the maximum height, the final velocity is 0.

Form first equation.

$\rm 0=49-32.2t$

On simplifying, we have

$\rm t = 1.52 \ sec$

b) The highest point will be.

Then from the third equation.

At the maximum height, the final velocity is 0.

$\rm 0^2 = 49^2 - 2*32.2*h$

on simplifying, we have

$\rm h = 37.28$ ft

c)  The range of the function will be from 0 to 37.28 ft.

d) A maximum point on the graph of this function is 37.28 ft.

More about the projectile link is given below.

https://brainly.com/question/11261462