A toy rocket is launched from the top of a building 98 feet tall at an initial velocity of 194 feet per second.

Question

A toy rocket is launched from the top of a building 98 feet tall at an initial velocity of 194 feet per second.

​a) Give the function that describes the height of the rocket in terms of time t.

​b) Determine the time at which the rocket reaches its maximum​ height, and the maximum height in feet.

​c) For what time interval will the rocket be more than 462 feet above ground​ level?

​d) After how many seconds will it hit the​ ground?

Answer 1

h(t) = -16t^2 + 194t + 98

h'(t) = -32t +194 = 0

t + 194/32 = 6.0625 seconds to reach maximum height = about 6 seconds

h(6) = -16(6)^2 + 194(6) + 98 = 588 + 98 = about 686 feet max height

h(194/32) = -16(194/32)^2 + 194(194/32) + 98

= (-1/64^2)(37,636) +(2/64)(37636) + 98

= 37636/64 + 98

= 588.0625 + 98

= 686.0625 feet maximum height = about 686 feet

it hits ground when h(t) = 0

t= about 12.61 seconds, a little longer than twice time to reach max height because it started above ground

interval when h> 462 is from t = 2.21 to 9.81 seconds

calculate t from setting h(t) = 462 it's a quadratic equation, so there's two solutions

about 6.06 + and - 3.75 = 2.21 and 9.81 seconds

that time interval is twice 3.75 = 7.5 seconds it's above 462 feet high