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Question

An object is thrown upward at a speed of 50 feet per second by a machine from a height of 11 feet off the ground. The height  of the object after t seconds can be found using the equation s(t)=−16t^2+v0t+s0, where v0 is the initial velocity and s0 is the initial height. Give all numerical answers to 2 decimal places.

(a) When will the object reach its maximum height?   seconds.

(b) What is its maximum height?   feet.

Answer 1

You have to use s(t) equation and plug in the known quantities. For instance, you know the initial velocity (v0) which is 50 f/s and you know the initial height s0 which is 11 ft. If we replace what we know we now have

s(t) = -16 t^2 + 50t + 11

In order to find the maximum height we focus on finding the vertex of a parabola which is using (-b/2a) where a is the coefficient in front of the t^2 term and b is the coefficient in front of the t term. In doing that we have

-50 / 2 (-16) or 50/32 or 25/16 seconds. In decimal form that is 1.56 seconds.

For part b you plug in 25/16 into t and solve for s(t). Do not use your rounded answer of 1.56 otherwise your answer will be off.

-16 (25/16)^2 + 50(25/16) + 11 = 50.0625 or 50.06 feet.