Two soccer players, mary and jane, begin running from approximately the same point at the same time. mary runs in an easterly di
rection at 3.59 m/s, while jane takes off in a direction 42.8 ◦ north of east at 4.67 m/s. what is the magnitude of the velocity of jane relative to mary?
1 answer:
As you illustrate the problem, you may come up with the same diagram as the one shown in the attached picture. The relative velocity is the linear velocity between the two velocities. This is calculated by simply finding the difference of their velocities. Since, the velocity is with respect to Jane relative to Mary, the solution is as follows:
Relative velocity = Velocity of Jane - Velocity of Mary
Relative Velocity = 4.67 m/s - 3.59 m/s = 1.08 m/s
Relative velocity = Velocity of Jane - Velocity of Mary
Relative Velocity = 4.67 m/s - 3.59 m/s = 1.08 m/s
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Answer:
The maximum height reached is 4.63 m.
Explanation:
Given:
Initial speed of the man (u) = 31.0 m/s
Speed at the top of trajectory () = 29.5 m/s
Acceleration due to gravity (g) = 9.8 m/s²
When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.
Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.
So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.
Now, initial velocity can be rewritten in terms of its components as:
Where, are the initial horizontal and vertical velocities of the man.
Now, plug in the given values and simplify. This gives,
Now, we know that, for a projectile motion, the maximum height is given as:
Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,
Therefore, the maximum height reached is 4.63 m.