A sprinter leaves the starting blocks with an acceleration of 4.5m/s^2. What is the sprinter's speed 2 s later
2 answers:
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The motion of the ball on the vertical axis is an accelerated motion, with acceleration
The following relationship holds for an uniformly accelerated motion:
where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
So we can rewrite the equation as
from which we can isolate h
(1)
Now let's assume that
is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 
. So the maximum height of the second ball is
(2)
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
The following relationship holds for an uniformly accelerated motion:
where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
So we can rewrite the equation as
from which we can isolate h
Now let's assume that
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
Answer:
in the direction of motion of Jacob
Explanation:
Given:
mass of Jacob,
velocity of Jacob,
mass of Ethan,
velocity of Ethan,
Now using the conservation of linear momentum for the case:
(When the two masses in motion combine to form one after the collision then they will move together in the direction of the greater momentum.)
in the direction of motion of Jacob as it was assumed to be positive.