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

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.

The answer should be 5.8 meters :/

oops...

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

No :c really

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

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

Yes it is

I know that

A= 9.80

Velocity= 14

Time = 0.50

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

did you learn that F=ma ?

I didn't understand the formula

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

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

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.

Are you ok with the answer ?

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

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

Good Luck

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