(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
<u>The given parameters include:</u>
- constant velocity of the elevator, u₁ = 10 m/s
- initial velocity of the ball, u₂ = 20 m/s
- height of the boy above the elevator floor, h₁ = 2 m
- height of the elevator above the ground, h₂ = 28 m
To calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.
The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.
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Answer:
A box with a mass of 50 kg is raised straight up. What is the force of the box? plss help!! two types exist-positive and negative. possible answers: A- electric current B- repel C- lines of force D- charges. Five wavelengths are generated every 0.100 s in a tank of water.
Explanation:
I hope it helped
It would be Atoms, they’re all made up of these tiny particles
<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.
First, the vertical component of tension (Tsin theta) is equal to the weight of the object.
T * sin θ = mg =</span> 1.55 * 9.81 <span>
T * sin θ = 15.2055
Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.
T * cos θ = 13.3
Tan θ = sin </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82.
T then is equal to 20.20 N
At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is
∑ F = ma
n - 430 N = (430 N)/g • a
where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is
a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²
and so
n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N
