Rock thrown in air

[michaelcaba]

This one stumps me. Can anyone show me their steps to the solution, which is 2.5 seconds.

A rock is thrown straight up into the air from a height of 4 feet. The height of the rock above the ground in feet, t seconds after it is thrown is given by −16t^2+56t+4. For how many seconds will the height of the rock be at least 28 feet above the ground?

 

[Deleted member 4993]

This one stumps me. Can anyone show me their steps to the solution, which is 2.5 seconds.

A rock is thrown straight up into the air from a height of 4 feet. The height of the rock above the ground in feet, t seconds after it is thrown is given by −16t^2+56t+4. For how many seconds will the height of the rock be at least 28 feet above the ground?

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Please share your work/thoughts about this assignment.

Do you know how to solve quadratic equation?

 

[michaelcaba]

OK. I know that it takes 0.5 seconds to reach 28 feet. I solve this by solving the quadratic equation by factoring. Easy. But what stumps me is the part about how long it is at least 28 feet. Any hints, ideas? What am I missing?

 

[michaelcaba]

Oops. My bad. I figured it out. Never mind.