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
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:
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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