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An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = -4.9t2 +19.6t + 58.8, where s is in meters. How high will the object be after 3 seconds?
A)73.5 feet
B)98.49 feet
C)333.69 feet
D)161. 7 feet

Respuesta :

xero099

The ball will be 73.5 meters above the ground after a time of 3 seconds. (Correct choice: A)

What is the height of the object after launch?

Herein we have an equation that models the height of the ball as function of time, we need to evaluate the expression for t = 3 s to find the required height:

s(3) = - 4.9 · (3 s)² + 19.6 · (3 s) + 58.8

s(3) = 73.5 m

The ball will be 73.5 meters above the ground after a time of 3 seconds. (Correct choice: A)

To learn more on quadratic equations: https://brainly.com/question/2263981

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