Thread: [SOLVED] Initial velocity using Calculus not physics??

I have the equation x=ct-4.9t^2 where x=height, t=time for a body projected vertically up from the ground.

I want to find the initial velocity if the body just reaches a height of 250m using calculus (I already found the answer with Physics-see below).

The maths
x=ct-4.9t^2
x`=c-9.8t, c=9.8t
x``=-9.8

Can someone help?


s=(final velocity^2-initial velocity^2)/2A
250= (0^2-x^2)/2(-9.8)
X=70m/s

Originally Posted by Neverquit

I have the equation x=ct-4.9t^2 where x=height, t=time for a body projected vertically up from the ground.

I want to find the initial velocity if the body just reaches a height of 250m using calculus (I already found the answer with Physics-see below).

The maths
x=ct-4.9t^2
x`=c-9.8t, c=9.8t
x``=-9.8

Can someone help?


s=(final velocity^2-initial velocity^2)/2A
250= (0^2-x^2)/2(-9.8)
X=70m/s

"Using physics rather than calculus" strikes me as an odd expression since the physics equations are derived from calculus, but I understand your meaning.

At the peak of its trajectory, dx/dt = 0, so we set dx/dt = c - 9.8t = 0

Solve for t, to get t = c/9.8

This is the time at which x_max is reached

plug back into equation for x(t)

x_max = ct_max - 4.9t_max^2

250 = c(c/9.8) - 4.9(c/9.8)^2

...

c = 70 m/s