SOLUTION: If an object is projected upward with an initial velocity of 48 ft pe second from a height h of 160 ft, then its height t second after it is projected is defined by the equation h=

Question 200633: If an object is projected upward with an initial velocity of 48 ft pe second from a height h of 160 ft, then its height t second after it is projected is defined by the equation h= -16t 2 + 48t + 160. How many seconds after it is projected will it hit the ground?
Found 2 solutions by ankor@dixie-net.com, Earlsdon:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
If an object is projected upward with an initial velocity of 48 ft per second
from a height h of 160 ft, then its height t second after it is projected is
defined by the equation h= -16t 2 + 48t + 160.
:
How many seconds after it is projected will it hit the ground?
;
When the object strikes the ground h = 0, therefore:
-16t^2 + 48t + 160 = 0
:
Simplify, divide by -16
t^2 - 3t - 10 = 0
Factor
(t - 5)(t + 2) = 0
Positive solution
t = 5 seconds to hit the ground
:
:
Check solution in original equation
h = -16t^2 + 48t + 160
h = -16(5^2) + 48(5) + 160
h = -400 + 240 + 160
h = 0
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
You are looking for the time, t, when h(t) = 0, so...
Substitute h(t) = 0.
Rewrite in standard form:
First factor -16.
Notice the change of sign in each term in the parentheses as a result of factoring the -16. Now apply the zero product rule.
Solve by factoring.
Apply the zero product rule.
or so then...
or Discard the negative solution as the time, t, should be a positive value.
seconds.
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