How far does a ca4 go
1 answer:
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(a) 0.40 s
First of all, let's find the initial speed at which Jordan jumps from the ground.
The maximum height is h = 1.35 m. We can use the following equation:
where
v = 0 is the velocity at the maximum height
u is the initial velocity
is the acceleration of gravity
Solving for u,
The time needed to reach the maximum height can now be found by using the equation
Solving for t,
Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of
h' = 1.35 - 0.20 = 1.15 m
Using again the equation
we find
And the corresponding time is
So the time to go from h' to h is
And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:
(b) 0.08 s
This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:
And the corresponding time is
So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2: