SOLUTION: Suppose you shoot an arrow straight up at a speed of 60 miles per hour when it leaves the bow. From the laws of physics it is possible to deduce that t seconds after leaving the b

Question 88682: Suppose you shoot an arrow straight up at a speed of 60 miles per hour when it leaves the bow. From the laws of physics it is possible to deduce that t seconds after leaving the bow the height of the arrow from the ground (in feet) is:

h = -16t2 + 88t + 7

a. How fast is 60 miles per hour in feet per second?
b. How high is the arrow 2 seconds after being shot?
c. About how high does the arrow go before it starts falling back to earth?
d. When does it hit the ground?

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Suppose you shoot an arrow straight up at a speed of 60 miles per hour when it leaves the bow. From the laws of physics it is possible to deduce that t seconds after leaving the bow the height of the arrow from the ground (in feet) is:
:
h = -16t^2 + 88t + 7
:
a. How fast is 60 miles per hour in feet per second?
60 mph is 1 mi per minute
Then 5280/60 = 88 ft/sec
:
b. How high is the arrow 2 seconds after being shot?
Substitute 2 for t:
-16(2^2) + 88(2) + 7 =
-16(4) + 176 + 7
-64 + 176 + 7 = 119 ft
:
c. About how high does the arrow go before it starts falling back to earth?
Find the axis of symmetry: x = -b/(2a); in this equation: a=-16; b=88
t = -88/(2*-16)
t = -88/-32
t = +2.75 sec
Find the vertex, which is the max in this problem, substitute 2.75 for t
h = -16(-2.75^2) + 88(2.75) + 7
h = -16(7.5625) + 242 + 7
h = -121 + 242 + 7
h = 128 ft is max height, begins falling to earth from there (in 2.75 sec)
:
d. When does it hit the ground?
The height of the ground is 0 so we have:
-16t^2 + 88t + 7 = 0
:
Have to use the quadratic formula to find t:

:
Solve this you should get a positive solution of:
t ~ 5.58 sec to reach the ground

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