Suppose u shoot a arrow into the air from 320 ft high building with an initial upward velocity of 128 feet/second. what is the arrows height after 3 seconds? What is the maximum height the arrow obtains before falling back down towards the ground? When does it reach its maximum height f(t) = -16t+128+320
hi!!
When is the arrow 100 ft above the ground?
Helli
think you copied this incorrectly right? \\[f(t)=-16t^2+158t+320\]
what is the arrows height after 3 seconds? put \[f(3)=-16\times 3^2+128\times 3+320\] to find your answer
What is the maximum height the arrow obtains before falling back down towards the ground? the max is when \(t=-\frac{b}{2a}=-\frac{128}{2\times (-16)}=4\) plug in \(4\) to get the maximum height
What is the maximum height the arrow obtains before falling back down towards the ground? set \[-16t^2+128t=320=100\] or \[-16t^2+128t+220=0\] and solve the quadratic equation for \(t\)
Ohhhhh OK I thought I did it wrong thank u so much
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