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
Don't hesitate to provide me a nice feedback. I will appreciate. Thanks.
The answer should be 5.8 meters :/
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
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
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
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
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 :)
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