A projectile launcher upward with a velocity of 160 feet per second from the top of a 70-foot platform. What is the maximum heig
ht attained by the projectile?
1 answer:
Answer:
470 ft
Step-by-step explanation:
The equation for ballistic motion is usually written ...
h(t) = -16t^2 +v0·t +h0
where v0 is the initial upward velocity in ft/s, and h0 is the initial height in ft.
Filling in the numbers given in the problem statement, the equation becomes ...
h(t) = -16t^2 +160t +70
This can be rewritten to vertex form as follows.
h(t) = -16(t^2 -10t) +70
= -16(t^2 -10t +25) +70 +400 . . . . . . add and subtract 400 to complete the square
= -16(t -5)^2 +470
The vertex of the height vs. time curve is at (t, h) = (5, 470).
The maximum height attained is 470 feet.
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Answer:
i) 1350 m
ii) 5400 m
Step-by-step explanation:
<u>Formula for Speed</u>
Rearrange the formula so that Distance is the subject:
Given:
- Speed = 45 m/s (meters per second)
- Time = 30 seconds
Substitute the given values into the formula for distance:
Given:
- Speed = 45 m/s (meters per second)
- Time = 2 minutes
As the time is given in a different unit of time as the speed, we must first convert the time into seconds:
1 minute = 60 seconds
⇒ 2 minutes = 60 × 2 = 120 seconds
Therefore:
- Speed = 45 m/s (meters per second)
- Time = 120 seconds
Substitute the values into the formula for distance: