Question

A steel ball is dropped onto a hard floor from a height of 1.95 m and rebounds to a height of 1.89m. (Assume that the positive direction is upward.) 

(a) Calculate its velocity (in m/s) just before it strikes the floor 

(b) Calculate its velocity (in m/s) just after it leaves the floor on its way back up

(c) Calculate its acceleration (in m/s) during contact with the floor if that contact lasts 0.0800 ms 

(d) How much (in ) did the ball compress during its collision with the floor assuming the floor is absolutely rigid? 

An athlete is training on a 100 m long linear track. His motion is described by the graph of his position vs. time, below.

(a) For each segment of the graph, find the magnitude and direction of the athlete's velocity

 (b) What are the magnitude and direction of the athlete's average velocity over the entire 60 s interval? 

A particle is restricted to move along one dimension, the x-axis. The graph below plots the velocity of the particle as a function of time.

 (a) What is the acceleration of the particle during the following intervals? Indicate the direction with the sign of your answer.

 (b) At the following Instants in time, what is the direction of the particle's velocity?

Answer 1

the height from which the ball is dropped , h1 = 1.95 m

the height to which the ball rebounds , h2 = 1.89 m

a)

the velocity of ball just before it strikes the floor , u = - sqrt(2*g*h1)

u = - sqrt(2*9.81*1.95) = 6.19 m/s

b)

the velocity of ball just after it leaves the floor , v = sqrt(2*g*h1)

v = sqrt(2*9.81*1.89) = 6.09 m/s

c)

time taken , t = 0.08 ms = 8 * 10^-5 s

let the acceleration be a

v =u + a * t

6.09 = - 6.19 + a * 8 * 10^-5

a = 1.53 * 10^5 m/s^2

d)

let the compression in the ball be s

using third equation of motion

v^2 - u^2 = 2 * a * s

6.09^2 - 6.19^2 = 2 * 8 * 10^5 * s

s = 7.68 * 10^-7 m