A traffic accident detective measures the skid marks left by a car, 825kg. He determines that the distance between the point tha
1 answer:
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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.
Answer:
The tension in each half of the rope, is approximately 4,908.8 N
Explanation:
The mass of the acrobat, m = 60 kg
The length of the rope, l = 10 m
The extent by which the center dropped = 30 cm = 0.3 m
Let, 'T' represent the tension in each half of the rope
Weight, W = Mass, m × The acceleration due to gravity, g
∴ W = m × g
The acceleration due to gravity, g ≈ 9.8 m/s²
∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N
The angle the dropped rope makes with the horizontal, θ is given as follows;
θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°
At equilibrium, the sum of vertical forces, = 0
The vertical component of the tension, , in each half of the rope is given as follows;
= T × sin(θ)
∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)
Plugging in the values, with θ = arctan(0.06) for accuracy, we get;
588 N + 2 × T × sin(arctan(0.06) = 0
∴ 2 × -T × sin(arctan(0.06) = 588 N
-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N
The tension in each half of the rope, T ≈ 4,908.8 N.
