SOLUTION: An object is launched into the air from a ledge 6 feet off the ground at initial vertical velocity of 96 feet per second. Its height H, in feet, at t second is given by the equatio
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-> SOLUTION: An object is launched into the air from a ledge 6 feet off the ground at initial vertical velocity of 96 feet per second. Its height H, in feet, at t second is given by the equatio
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Question 1087720: An object is launched into the air from a ledge 6 feet off the ground at initial vertical velocity of 96 feet per second. Its height H, in feet, at t second is given by the equation H = 16t^2+96t+16. Find all the times t that the object is at a height of 160 feet off the ground. Found 2 solutions by ikleyn, josmiceli:Answer by ikleyn(52778) (Show Source):
You can put this solution on YOUR website! The equation should be:
( I noted you said at the
object was ft off the ground,
it should be ft. )
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The object is at a height of 160 ft
only once, at t = 3 sec
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check:
here's the plot:
My answer looks about right
