Vertical Ball Movement: Calculating Time in Flight (Calculus Practice)

[VikingStorm]

[SOLVED] Integration (story)

Use a(t) = -32 feet per second squared as the acceleration due to gravity. A ball is thrown vertically upward from the ground with an initial velocity of 56 feet per second. For how many seconds will the ball be going upward?

(Calculus question)

I don't see where I would pull in calculus into this. (Will be on Calc test tomorrow, so other tips to dissect these kind of problems would be nice)

 

[Hurkyl]

Well, what do you know that you can apply to this?

(P.S. the night before is generally a little late to be studying for a test; you'd probably be better off getting a good night's rest)

 

[M98Ranger]

Hi there,

In order to solve this problem using calculus, we can use the equation d(t) = -16t^2 + v0t + d0, where d(t) is the height of the ball at time t, v0 is the initial velocity, and d0 is the initial height (in this case, 0). We can use this equation to find the time at which the ball reaches its maximum height, which would be the time at which it stops going upward and starts falling back down.

To find this time, we can take the derivative of the equation with respect to time, which gives us v(t) = -32t + v0. Setting this equal to 0 and solving for t, we get t = v0/32. Plugging in the given initial velocity of 56 feet per second, we get t = 56/32 = 1.75 seconds.

Therefore, the ball will be going upward for 1.75 seconds before it starts falling back down. I hope this helps and good luck on your calculus test tomorrow! Remember to always break down the problem into smaller parts and use the appropriate equations and techniques.