Let the key is free falling, therefore from equation of motion
.
Take initial velocity, u=0, so
.
As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula
From above substituting t,
.
Now substituting all the given values and g = 9.8 m/s^2, we get
.
Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.
Answer:
0.074m/s
Explanation:
We need the formula for conservation of momentum in a collision, this equation is given by,
Where,
= mass of ball
= mass of the person
= Velocity of ball before collision
= Velocity of the person before collision
= velocity of ball afer collision
= velocity of the person after collision
We know that after the collision, as the person as the ball have both the same velocity, then,
Re-arrenge to find ,
Our values are,
= 0.425kg
= 12m/s
= 68.5kg
= 0m/s
Substituting,
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<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>