Question

A rocket project straight up with a launch velocity of 250m/s. It reaches a height of s = 250t - 100t² after t second. How high does the rocket go? (Round your answer to 2 decimal places)​

Answer 1

The rocket reaches a maximum height of approximately 250 * t - 100 * t^2, where t is the time in seconds. To find the maximum height, we need to find the value of t when the derivative of the height function, s = 250t - 100t², is equal to zero. We can find the derivative by the power rule and get s' = 250 - 200t. Setting s' = 0, we get:

250 - 200t = 0

t = 250/200 = 1.25

Plugging this value of t into the original equation for height, we get:

s = 250 * 1.25 - 100 * (1.25)^2 = 312.50 - 156.25 = 156.25 m.

So the rocket goes 156.25 m high.

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