s(t)=-16t^2+vot An object is projected vertically upward with an initialvelocity

Question

s(t)=-16t^2+vot An object is projected vertically upward with an initialvelocity of vo ft/sec, and its distance s(t) in feet above theground after t seconds is given by the equation above. A) If the object hits the ground after 12 seconds, findits initial velocity vo B) Find its maximum distance above the ground. s(t)=-16t^2+vot An object is projected vertically upward with an initialvelocity of vo ft/sec, and its distance s(t) in feet above theground after t seconds is given by the equation above. A) If the object hits the ground after 12 seconds, findits initial velocity vo B) Find its maximum distance above the ground.

Explanation / Answer

So 0 = -16t^2+vot = -16(144)+12vo vo = 192ft/s
B) Your new equation is s(t) = -16 t^2 + 192t i) Find the derivative of this equation. s'(t) = -32 t + 192
ii) Set this equal to 0. When the derivative is 0, you have amaximum or minimum. In this case, a maximum 0 = -32 t +192 t = 6
iii) Find the height at t =6 from the original equation s(t) = -16 (6)^2 +192 (6) s(t) = 576 feet

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