math
posted by cris .
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
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 800t16t^2 = 0
Solve for t in 80016t^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
thank you Steve for your help
Respond to this Question
Similar Questions

Physics
A bullet is dropped into a river from a very high bridge. At the same time, another bullet is fired from a gun straight down towards the water. If air resistance is negligible, the acceleration of the bullets just before they strike … 
math
If an object is fired upward from the top of a tower at a velocity of 80 feet per second, the tower is 200 ft high, the formula is h(t)=16t^2+80t+200, t is time h is height, how long after it is fired does the object reach the max … 
Physics
A ball is thrown straight up in the air and passes a window 0.30s after being released. It takes 1.5s to go from the window to its maximum height and back down to the window. What was the initial velocity of the ball when it was released? 
physics
a cannonball is fired horizontally from the top of a cliff. the cannon is at height H=60.0m above ground level , and the ball is fired with initial horizontal speed v_0 .assume acceleration due to gravity to be g=9.80 find d/2 int … 
math
using this information: if an object is thrown straight up into the air from the height H feet per second then at time tseconds the height of the object is 16.1t^2 +Vt+ H feet. This formula uses only gravitational force, ignoring … 
math
using this information: if an object is thrown straight up into the air from the height H feet per second then at time tseconds the height of the object is 16.1t^2 +Vt+ H feet. This formula uses only gravitational force, ignoring … 
math. can someone explain this!
using this information: if an object is thrown straight up into the air from the height H feet per second then at time tseconds the height of the object is 16.1t^2 +Vt+ H feet. This formula uses only gravitational force, ignoring … 
Math
If an object is launched straight up into the air from a starting height of h_{0} feet, then the height of the object after t seconds is approximately h=16t^2+v_{0}t+h_{0} feet, where v_{0} is the initial velocity of the object. Find … 
math
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^2v_0 t Suppose the object is fired straight upwards with … 
Calculus the derivative
The height in feet of a free falling object t seconds after release is s(t)=16t^2+ v_0t+s_0, where s_0 is the height(in feet) at which the object is realsed, and v_0 is the initial velocity (in feet per second). Suppose the coin is …