A model rocket is launched straight upward from the side of a 212-ft cliff. The

Question

A model rocket is launched straight upward from the side of a 212-ft cliff. The initial velocity is 86 ft/sec. The height of the rocket h(t) is given by: h(t)= - 16t^2 + 86t + 212 where h(t) is measured in feet and t is the time in seconds. Use the quadratic formula to find the times (t) at which the rocket is 320 ft. above the ground. Round your answers to the nearest hundredth of a second. This is a 10-point question so make sure you are showing ALL steps. You will not be granted a redo if your work is lacking.

Explanation / Answer

Solution

The roots (solutions) of the quadratic equation, ax2 + bx + c = 0. are given by

x = {- b ± (b2 – 4ac)}/2a.

In the given equation on height, if height, h is substituted with 320 and then solved for t. the required answer will be obtained.

So, we have 320 = - 16t2 + 86t + 212 or 16t2 - 86t + 108 = 0 or 8t2 - 43t + 54 = 0.

This is a quadratic in t with a = 8, b = - 43 and c = 54.

Substituting these values in the formula given above,

t = {43 ± (432 – 4x8x54)}/(2x8) = {43 ± (1849 – 1728)}/(16) = (43 ± 11)/16 = 32/16 and 54/16 = 2 and 3.375.

Thus, the rocket will attain a height of 320 ft at time 2 seconds and 3.38 seconds ANSWER       

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