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A projectile launcher upward with a velocity of 160 feet per second from the top of a 70-foot platform. What is the maximum height attained by the projectile?

User Aeldron
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1 Answer

3 votes

Answer:

470 ft

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.

A projectile launcher upward with a velocity of 160 feet per second from the top of-example-1
User Eugenevd
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