3. An astronaut on a space-walk swings a 0.5 kg metal tool in a circular path at the end of a cable 3.6 m long. The astronaut th
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speed of tortoise is given as v1 = 0.14 m/s
speed of hare is given as v2 = 20*0.14 = 2.8 m/s
now let say the total length of the path is "d"
so the total time taken by the tortoise to cover this
now given that hare took rest for 1 min
so total time of run for hare is (t - 60)s
so the distance that hare covered is given by
now by above two equations
and the time t is given by
so part a)
t = 63 s
part b)
d = 8.82 m
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.