Question

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

Answer 1

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.

...

Answer 2

thank you Steve for your help