# Differentials help word problem?? (i think you need differentials!)

• Nov 30th 2010, 11:43 PM
yess
Differentials help word problem?? (i think you need differentials!)
The gravity of earth is the constant 9.8m/s^2. when an object is propelled upwards gravity works as negative acceleration. Air resistance is ignored
a) find v(t), velocity in terms of time, of an object propelled straight up with an initial velocity (at t=0) of 20 m/s
b) Given v(t), find the height function for the object given that the object begins its upward trajectory at a height of 1.5m
c) Find the maximum height reached by the object

i just have noooo idea what to do here!
• Nov 30th 2010, 11:53 PM
Mollier
Hi mate, I'll get you started.

Basically your velocity decreases as a function of time.
At the beginning, \$\displaystyle v(t_0)\$, your velocity is \$\displaystyle 20m/s\$.
Each second, it decreases \$\displaystyle 9.8m/s\$ because the gravity pull is \$\displaystyle 9.8m/s^2.\$

That means that the velocity as a function of time is,

v\$\displaystyle (t) = v(t_0) - at = 20m/s - 9.8m/s^2t\$.

Can you do the others now?
• Dec 1st 2010, 02:37 PM
yess
Quote:

Originally Posted by Mollier
Hi mate, I'll get you started.

Basically your velocity decreases as a function of time.
At the beginning, \$\displaystyle v(t_0)\$, your velocity is \$\displaystyle 20m/s\$.
Each second, it decreases \$\displaystyle 9.8m/s\$ because the gravity pull is \$\displaystyle 9.8m/s^2.\$

That means that the velocity as a function of time is,

v\$\displaystyle (t) = v(t_0) - at = 20m/s - 9.8m/s^2t\$.

Can you do the others now?

so would h(t) = h(t0) - at so 1.5m - 9.8m/s^2t ?? also what about all the varying m/s and m/s^2 ? do we just disregard those so it would be like 20-9.8t for velocity?