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) = - 4.9t^2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground?

Question

Mathematics

Victor Stevenson

28 November, 06:22

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) = - 4.9t^2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground?

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Answers (1)

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Tyree Bradley · Answer 1

So the question sets the problem with a bunch of different values given but it goes ahead does the work of putting them into an equation, since this equation they made gives us the position, when it is equal to zero the object has impacted the ground. Setting it to zero allows us to solve for t, the time at which it strikes the ground.

0 = - 4.9t^2 + 19.6t + 58.8, because this is a quadratic, and a complicated one at that, we will use either a calculator, or the quadratic formula to solve for t.

No matter what way you solve it you will get two values, only the positive value applies because a negative value for time is illogical.

The answer is at t = 6 seconds