Posted by **cris** on Sunday, May 25, 2014 at 3:09pm.

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 -
**MathMate**, Sunday, May 25, 2014 at 3:17pm
You may want to check the equation to see if there is a typo. I expect it to read:

h=-16t²+v_0 t

since the object is thrown straight *up*wards, while gravity (downwards) is indicated negative.

If it is fired straight upwards at an initial speed of 800 ft/s, the initial velocity is +800 ft/s, i.e. upwards.

- math -
**cris**, Tuesday, May 27, 2014 at 2:27pm
yes, you're right. the equation is h=-16t^2+v_0t

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