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A toy rocket is launched vertically upward from ground level with an initial velocity of 128 ft/s.

1. How long will it take for the rocket to return to the ground?
2. When is the rocket 32 feet above ground?

Answer :

Note: As you missed to mention the given equation for t seconds and height h, so I am taking a sample equation h(t) =-16t² + 28t + 40. So, I am explaining your question based on this equation, which would anyways clear your query.

Answer:

It will take 2 seconds for the rocket to return to the ground when is the rocket 32 feet above ground.

Note: Sample equation h(t) =-16t² + 28t + 40 was used to solve this problem, as you had not mentioned the equation.

Step-by-step explanation:

To determine:

How long will it take for the rocket to return to the ground when is the rocket 32 feet above ground?

Information Fetching and solution steps:

  • Initial Velocity = 28 ft/s
  • The equation for height h and second t is h(t) = -16t² + 28t +40

So,

Let us consider the equation h(t) = -16t² + 28t + 40

32 = -16t² + 28t + 40

To find out how long will it take for the rocket to return to the ground when is the rocket 32 feet above ground, plug in h(t) = 32ft, rearrange into quadratic form, and solve:

32 = -16t² + 28t + 40

0 = -16t² + 28t + 8

Step 1: Factor right side of equation

0 = −4(4t + 1)(t − 2)

−4(4t + 1)(t − 2) = 0

Step 2: Set factors equal to 0

4t + 1 = 0 or t − 2 = 0

t = -1/4 or t = 2

As t can not be negative, so t = 2 seconds.

Hence, it will take 2 seconds for the rocket to return to the ground when is the rocket 32 feet above ground.

Keywords: time, height, velocity


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Rewritten by : Barada

Answer:

1.14 s

Step-by-step explanation:

Time, [tex]t=\frac {d}{s}[/tex]

Here, d is the distance and s is the speed/velocity

Since we're given the velocity, s as 28 ft/s and the distance between the position of the rocket and ground as 32 ft then

[tex]t=\frac {32}{28}=1.142857143\approx 1.14 s[/tex]

Therefore, it needs 1.14 seconds