The elastic ball bounces down a flight of rigid stairs. Coefficient of restitution is e = 0.85 and the initial velocity of the ball is 6 Neglecting friction between sec the ball and stairs, which of the following most nearly gives the value of the height h of the steps if all the trajectories between successive steps are identical?

The elastic ball bounces down a flight of rigid stairs. Coefficient of restitution is e = 0.85 and the initial velocity of the ball is 6 Neglecting friction between sec the ball and stairs, which of the following most nearly gives the value of the height h of the steps if all the trajectories between successive steps are identical?

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

Gayle runs at a speed of 4.10 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 5.00 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 15.0 m? Gayle's mass is 46.5 kg, the sled has a mass of 5.25 kg and her brother has a mass of 30.0 kg.
A 10 kg block A rests on a smooth surface and the spring is compressed 0.1 m. It is attached to an inextensible cord over frictionless, weightless pulleys with a mass B of kg, as shown. If the system is released from rest and B travels 0.1 m, what is A's speed? Consider K = 490N/m (spring) and V₁ = 2 x VB
Determine the coefficient of restitution e for a steel ball dropped from rest at a height h above a heavy horizontal steel plate if the height of the second rebound is h2. Can someone please explain to me how the book is getting that equation?
For a perfectly elastic collision, u1+v1=u2+v2, or alternatively, u2−u1=v1−v2. If object 2 is initially at rest, then u2=0. Then v1−v2=−u1. However, for a partially elastic collision the relative velocity after the collision will have a smaller magnitude than the relative velocity before the collision. We can express this mathematically as v1−v2=−ru1, where r (a number less than one) is called the coefficient of restitution. For some kinds of bodies, the coefficient r is a constant, independent of v1 and v2. Show that in this case the final kinetic energy of the motion relative to the center of mass is less than the initial kinetic energy of this motion by a factor of r2. Furthermore, derive expressions for v1 and v2 in terms of u1 and r.
On a horizontal frictionless table, a 3-kg block moving 5 m/s to the right collides with an 8-kg block moving 1.5 m/s to the left. b. If the two blocks stick together, what is the final velocity (magnitude and direction)? a. How much mechanical energy is dissipated in the collision? Please show the given, formula, figure, and free body diagram.
5kg sphere A travels with velocity Va= 10m/s colliding with 3kg sphere B that is traveling at Vb= 10m/s. coefficient of restitution is 0.4, find magnitude of velocities of each sphere immediately after impact.
A package is dropped from a height h. A high-speed camera is used to determine that the rebound height is ¼ of the drop height. If the impact velocity is 176 in/s, find the drop height, the coefficient of restitution, and the velocity change during the impact.
A jogger with mass m = 55 kg uses energy at a rate of P = 802 W when she jogs on horizontal ground at a constant speed of s = 3.5 m/s. The efficiency of her “human engine” however, is e = 0.27. (In other words, the power used for locomotion is e = 0.27 times the total power.) Q) Write an expression for the rate of energy expenditure whenshe jogs at the same constant speed up a slope of 4.99 degrees. Assume whatever frictional or drag forces are the same in each case. Your answer should be in terms of the symbols given in the problem statement and g(9.80 m/s^2)
Forming a shock absorber, two springs of equal length are nested together. If the maximum compression of the springs is 0.25 and the outer spring has a stiffness of Ka = 420 N/m, what stiffness is required of the inner spring Kb? The design of the spring system is such that if a mass of 2.5 kg is at rest, then dropped from a distance of 0.75 m above the springs, the spring system will catch and stop the mass.
To apply Newton’s second law and the theorem of conservation of energy to solve kinetic problems. A bungee jumper wants to jump off the edge of a bridge that spans a river below. The jumper has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has lengthL when unstretched, will first straighten and then stretch as the jumper falls.Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch and has spring constant k. The jumper does not actually jump but simply steps off the edge of the bridge and falls straight downward. The jumper's height is negligible compared to the length of the bungee cord. Thus, the jumper can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity. How far below the bridge, d, will the jumper eventually be hanging, once the jumper stops oscillating and comes finally to rest? Assume that the jumper does not touch the water. Express your answer in…
A SMALL 0.65 BOX IS LAUNCHED FROM REST BY A HORIZONTAL SPRING .THE BLOCK SLIDES ON A TRACKDOWN A HILLAND COMES TO REST AT A DISTANCE d FROM THE BASE OF THE HILL .KINEIC FRICTION BETWEEN THE BOX AND THE TRACK IS NEGLIGIBLE ON THE HILL,BUT THE COEFFICIENT OF KINETIC FRICTION BETWEEN THE BOX AND THE HORIZONTAL PARTS OF TRACK IS 0.35.THE SPRING HAS A SPRING 34.5N AND IS COMPRESSED 30cm WITH THE BOX ATTACHED.THE BOX REMAINS ON THE TRACK AT ALL TIMES. (a)xcalculate the speed of the box before it starts descending down the hill
Wile E. Coyote runs into a painting of a tunnel at 15 mph at a 30◦ angle as shown below. After bouncing off of the painting with a coefficient of restitution e = 0.8 he slides along the ground with a coefficient of friction of μk = 0.4. Where does he come to rest?