A basketball is thrown straight up in the air. At its peak, it is 0.02 km high and has a velocity of 0 m/s. What is its final ve
1 answer:
You might be interested in
For counting x you use simple equation for the distance covered by the object when it moves with constant velocity:
that gives you 20m after 1st second, 40 m after 2nd second, 60 m after 3rd second and so on.
For counting y you have to use the equation for the distanced covered by the object moving with constantly accelerating velocity (symbols refering to vertical movement):
that gives you 5m after 1st second, 20m afters 2nd second, 45m after 3rd second and so on.
Add minus signs before y positions to receive graph presenting the movement of the ball.
So the points are: P1=[20,-5], P2=[40,-20], P3=[60,-45] and so on... Pn=[x,y].
that gives you 20m after 1st second, 40 m after 2nd second, 60 m after 3rd second and so on.
For counting y you have to use the equation for the distanced covered by the object moving with constantly accelerating velocity (symbols refering to vertical movement):
that gives you 5m after 1st second, 20m afters 2nd second, 45m after 3rd second and so on.
Add minus signs before y positions to receive graph presenting the movement of the ball.
So the points are: P1=[20,-5], P2=[40,-20], P3=[60,-45] and so on... Pn=[x,y].
To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.
By definition we know that the relative velocity of an object with reference to the Light, is defined by
Where,
V = Speed from relative point
c = Speed of light
On the other hand we have that the linear momentum is defined as
P = mv
Replacing the relative velocity equation here we have to
Therefore the height with respect the observer is
Therefore the height which the observerd measure for her is 0.56m