##### I have a physics test coming soon

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An object is thrown upwards with a speed of 14 m/s. How high above the projection point is it after 0.50 s

May 13th, 2015

you have the following relation :

speed=distance/time or distance = speed * time so the distance achieved above the projection point will be

14*0,5=7 meters

Don't hesitate to provide me a nice feedback. I will appreciate. Thanks.

May 13th, 2015

The answer should be  5.8 meters :/

May 13th, 2015

oops...

you are right. Do you know how to find this value ?

May 13th, 2015

No :c really

May 13th, 2015

I got this question from my practice midterm, it already has the answer but I don't know how to get it

May 13th, 2015

ok, we are going to work on it, first I tought It was only a simple mathematics calculus, but it's about physics

May 13th, 2015

Yes it is

May 13th, 2015

I know that

A= 9.80

Velocity= 14

Time = 0.50

May 13th, 2015

you have to find the equation of the trajectory of this object that we suppose are subjected to only the gravity

May 13th, 2015

did you learn that F=ma ?

May 13th, 2015

I didn't understand the formula

May 13th, 2015

Sum of powers practiced on an object is mass times acceleration known as 2nd Newton's law

May 13th, 2015

I believe you have not studied it yet so let's go for an easier way

May 13th, 2015

a(t)=-9.80

We put the minus because weight is oriented down

we integer to find v(t) the speed

v(t)=-9.80t+A where A is a constant and we know that v(0)=14 so we find that A=14

v(t)=-9.80t+14

we integer a last time to find the altitude of the object

z(t)=-1/2*9.80t^2+14t+B where B is also a constant and as z(0)=0 B = 0

os z(t)=-1/2*9.80t^2+14t

z(0,5)=5.8 meters.

That's it.

May 13th, 2015

Are you ok with the answer ?

May 13th, 2015

Yes thanks a lot! I have problems with understanding the formula  but I will study it thanks!

May 13th, 2015

You don't really need the formula to understand this problem :)

Good Luck

May 13th, 2015

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May 13th, 2015
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May 13th, 2015
Oct 19th, 2017
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