Question 775074: A man is standing over a beach and throws a rock up in the air from a height of 192ft. the initial velocity is 64 ft. per second. The rock height=h above the beach, in feet, after t seconds is given by the function
h(t)= -16t+64t+192
the -16t has an exponant of 2 but i could not figure out how to do it with my computer.
a) how high above the ground is the rock after 2 seconds?
b) How long will it take for the rock to raech the ground. o ft.?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A man is standing over a beach and throws a rock up in the air from a height of 192ft.
the initial velocity is 64 ft. per second.
The rock height=h above the beach, in feet, after t seconds is given by the function
h(t)= -16t^2 + 64t + 192; Use the "^", upper case 6
:
a) how high above the ground is the rock after 2 seconds?
Replace t with 2, solve for h(2)
h(2) = -16(2^2) + 64(2) + 192
h(2) = -64 + 128 + 192
h(2) = 256ft
:
b) How long will it take for the rock to raech the ground. o ft.?
The h= 0, so we have
-16t^2 + 64t + 192 = 0
Simplify, divide by -16
t^2 - 4t - 12 = 0
factors to
(t-6))(t+2) = 0
the positive solution is all we want here
t = 6 sec to reach ground
:
:
YOu can confirm this by finding h(t) when t = 6
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