SOLUTION: The path of a falling object is given by the function s= -16t^2 +v0t +s0 where v0 represnts the initial velocity in ft/sec and s represents the initial height. The variable t is t
Question 140648: The path of a falling object is given by the function s= -16t^2 +v0t +s0 where v0 represnts the initial velocity in ft/sec and s represents the initial height. The variable t is time in seconds, and s is the height of the object in feet. So if a rock is thrown upward with an initial velocity of 32 ft per second fron the top of a 40-ft building. How high is the rocke after 1 second?after 0.5 seconds? How many sec will it take the rock to reach maximum height before it's downward fall? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The path of a falling object is given by the function s= -16t^2 +v0t +s0 where v0 represents the initial velocity in ft/sec and s represents the initial height. The variable t is time in seconds, and s is the height of the object in feet. So if a rock is thrown upward with an initial velocity of 32 ft per second from the top of a 40-ft building. How high is the rock after 1 second?
:
The equation for this situation:
s = -16t^2 + 32t + 40
:
After 1 sec, substitute 1 for t:
s = -16(1^2) + 32(1) + 40
s = -16 + 32 + 40
s = 56 ft after 1 sec
:
after 0.5 seconds? Substitute .5 for t
s = -16(.5^2) + 32(.5) + 40
s = -16(.25) + 16 + 40
s = -4 + 16 + 40
s = 52 ft after .5 sec
:
How many sec will it take the rock to reach maximum height before it's downward fall?
The max will occur at the axis of symmetry; Use the formula: x = -b/(2a)
a = -16, b = 32
x =
x =
x = 1 sec
:
We found the height after one sec and it was 56 ft
