Time remaining:
##### where t is measured in seconds

label Algebra
account_circle Unassigned
schedule 0 Hours
account_balance_wallet \$5

where t is measured in seconds. If an object hits the ground after falling for 4 seconds, find the height from which the object was dropped. t= square root h/4.9

Oct 21st, 2017

hmm. I'm not sure I understand the equation all that well. Message me, I will show you how to do it.

Feb 16th, 2015

When an object is dropped to the ground from a height of h meters, the time it takes for the object to reach the ground is given by the equation t = Square root h/4.9, where t is measured in seconds. If an object hits the ground after falling for 4 seconds, find the height from which the object was dropped.

Feb 16th, 2015

Ok. Let's use calculus (because calculus and physics go hand in hand)

A derivative is a rate of change. Velocity is the rate of change of position. Acceleration is the rate of change of velocity.

An anti-derivative is the opposite of a rate of change. It's whatever is changing. So like, position is the anti-derivative of velocity. Velocity is the anti-derivative of acceleration.
I can teach you the rules of deriving or anti-deriving if you would like. You'll have to let me know. But for the purpose of this question I will just do it.

Acceleration due to gravity is -9.8 m/(s^2)

We need to find the velocity when it's at the highest point.
The equation for this is -9.8t+c (c is just some constant that get's canceled out when deriving).

V(t)=-9.8t+c        We know the velocity when t=0 is 0, because it's at the highest point and hasn't moved.

So let's solve for c. V(0)=0     so 0=-9.8(0)+c   0=c  c=vo (v knot, which means initial velocity, which just means the velocity when t=0.

So now we need the position function. I'll anti-derive the velocity function.

S(t)= ((-9.8t^2)/2)+vot+c

We don't know the position when t=0 yet, but we do know that at 0 seconds the object is at it's highest. So we can say that so (or s knot, which is just the initial position when t=0) is equal to s(0)=-4.9t^2+vot+c. We don't know c, but we can solve for it.

s(0)=So      So= -4.9(0)^2+vo(0)+c   So=c

So we know that the constant is equal to the initial position.
Which means we'll essentially be solving for so. If we were solving for t at this point we'd rearrange the equation to look like this:

Now we can solve for the position for when t=4. And find the initial position.

s(4)=-4.9(4)^2+vo(4)+so

0=-4.9(16)+0+so

0=-78.4+so= 78.4+so

So we know the height when we started was 78.4 meters off of the ground. 4 seconds after the object was dropped it hit the ground.

You can check by plugging this value back into your position equation. You now know so=78.4 meters.

so s(4)=-4.9(4)^2+78.4

-4.9(4)^2=-78.4

-78.4+78.4=0 which =s(4) (the position of the object after 4 seconds.)

I really hope this helps. Let me know if there's anything else I can do for. Take care partner.

Feb 16th, 2015

...
Oct 21st, 2017
...
Oct 21st, 2017
Oct 22nd, 2017
check_circle