# Differential Equations - Elementary Application - Vertical Motion

Mechanics Problem. I need it tomorr before 7 am. Thank you.

A ball is thrown upward with an initial velocity of 49 m/s. Find the velocity after 3 seconds. Find its maximum height. Compute the time it takes for the ball to return to its original point. Use g= 9.81 m/s^2 .

## Using the formula:

Using the formula:

a= dv/dt

By integration, I got:

v= at + vsub0

By integrating it again for the second time, wherein v= (ds/dt), I got:

s= 1/2at^2 + tVsub0 + Ssub0.

I'm quite confused. How can I get the value of my acceleration and where can I use the gravitational constant?

How can I compute the maximum height? And the time for the ball to return to its original point?

## Both of your equations are

Both of your equations are correct and to make your equation for distance s simpler, let your initial point be the origin so that your s

_{o}becomes zero.Your acceleration a = g. Take g as positive if your movement is downward and take g as negative if your movement is upward.

At its maximum height, your final velocity v = 0.

The time required to go up from the initial position to the highest point is equal to the time for the ball to fall from the highest point to its initial position.

## What formula can I use for

In reply to Both of your equations are by Jhun Vert

What formula can I use for computing the maximum height then? I tried solving it using physics formula H=[vi^2(sin^2theta)]/2g and I got 122.38. Which is wrong.. Correct answer is approx 5m.. How can I derive my formula using suvat equations?

## Use the formula s = 0.5at2 +

In reply to Both of your equations are by Jhun Vert

Use the formula s = 0.5at

^{2}+ v_{o}t but find first the value of t when v = 0 using the formula v = at + v_{o}. Note that your a = -g for upward motion and a = +g for downward motion.