Posted by **cris** on Thursday, May 29, 2014 at 10:17am.

An object is thrown or fired straight upwards at an initial speed of v_0 ft⁄s will reach height h feet after t seconds, where h and t are related to the formula

h=-16t^2+v_0 t

Suppose the object is fired straight upwards with an initial speed of 800ft⁄s, What is the initial velocity?

How does it change the equation h=-16t^2+v_0 t?

What is the initial position of the object?

When does the object fall back to the ground?

When does the object reach a height of 6400ft?

When does the object reach a height of 2mi?

How high is the highest point the ball reaches?

Suppose the object is dropped from a height of 288ft, what is v_0?

The equation becomes h=-16t^2+h_0 after (g) Why?

Write an equation which includes 288ft

- math -
**Steve**, Thursday, May 29, 2014 at 11:25am
initial velocity is 800 ft/s upward

it does not change the equation.

initial position is 0, since it was fired from the ground (height=0)

It falls back to the ground when h=0. So, solve for t in 800t-16t^2 = 0

Solve for t in 800-16t^2 = 6400

(why are there two solutions?)

Same as above, but you need to convert 2 miles to feet.

max height at the vertex of the parabola, when t is midway between the roots of the equation.

if dropped, v_0 is zero.

with v_0 = 0, the term vanishes, and we have an initial height, rather than initial speed.

...

- math -
**cris**, Friday, May 30, 2014 at 11:09am
thank you Steve for your help

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