Velocity problem/Max Height: If a particle is thrown....

[joshswimmer30]

Hi, im new to this site and am totally stumped by this problem.. Heres the question:

If a particle is thrown straight upward from the ground with an initial velocity of 160 ft/s, then its velocity after t seconds is v=-32t + 160 feet per second, and it attains a max height when t=5 s (and v=0). Compute the maximum height.

Im not really sure how to start, but i think the equation I should use is:

s=integral from a to b v(t)dt= x(b)-x(a)... possibly?

Thanks for the help guys

 

[joshswimmer30]

So if i do integral of -32t+160 and evaluate it from 0 to 5 will i get its max? or just its distance?

 

[galactus]

The derivative of the position function will give you the velocity function, therefore, the integral of the velocity function will give you the position function.

You're OK.

\(\displaystyle \L\\\int_{0}^{5}{-32t+160}dt\)

After you integrate and find the position function, graph it and see.

That always helps.

 

[TchrWill]

Re: Velocity problem/Max Height

Hi, im new to this site and am totally stumped by this problem.. Heres the question:

If a particle is thrown straight upward from the ground with an initial velocity of 160 ft/s, then its velocity after t seconds is v=-32t + 160 feet per second, and it attains a max height when t=5 s (and v=0). Compute the maximum height.

Im not really sure how to start, but i think the equation I should use is:

s=integral from a to b v(t)dt= x(b)-x(a)... possibly?

Maximum height derives from h = Vo(t) - 16t^2
Thus, h = 160(5) - 16(25) = 800 - 400 = 400 ft.