# math

posted by on .

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 - ,

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 upwards, 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 - ,

yes, you're right. the equation is h=-16t^2+v_0t

• math - ,

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