Question

To explore the physics of freefall, a curious student climbs to the top of the Las Vegas-version of the Leaning Tower of Pisa (to be built in the style of the Las Vegas-version of the Eiffel Tower, the Statue of Liberty, …) with two identical cannonballs and conducts a series of experiments. Part A) The student drops one cannonball, and exactly 1.0 s later drops the other cannonball from the same height. What is the time interval between the first cannonball striking the ground and the second cannonball striking the ground? Part B) The student drops one cannonball, and after it has fallen exactly 1.0 m drops the other cannonball from the same initial height. What is the distance between the cannonballs when the first cannonball strikes the ground? Part C) The student throws one cannonball directly upward at 5.0 m/s and simultaneously throws the other cannonball directly downward at 5.0 m/s. (This requires great dexterity, but the student has been practicing this maneuver for several months.) What is the difference in speed between the cannonballs when each ball strikes the ground?

Answer Choices: Less than 1.0 m 1.0 m Greater than 1.0 m Cannot be determined

Answer 1

a)

let the height travelled by each ball be "h"

for first ball dropped :

t = time of travel

a = acceleration = 9.8

v_o = initial velocity = 0 m/s

using the equation

h = v_o t + (0.5) a t²

h = 0 t + (0.5) (9.8) t²

t = sqrt(h/4.9)                                        eq-1

for the second-cannonbal dropped:

t' = time of travel = t + 1

using the equation

h = v_o t' + (0.5) a t'²

h = 0 t' + (0.5) (9.8) t'²

t' = sqrt(h/4.9)

using eq-1

t' = sqrt(h/4.9)

clearly time taken by each ball is same during the fall , hence the difference between interval of striking the ground is same as the delay of 1 sec

difference of time interval = 1 sec

b)

for the first cannonball :

velocity gained after travelling 1 m distance is given as

v_f² = v_o² + 2 a d

v_f² = 0² + 2 (9.8) (1)

v_f² = 19.6

Y = displacement = h -1

v_o = sqrt(19.6) m/s

t = time of travel to reach the ground

using the equation

Y = Y_o + v_o t + (0.5) a t²

Y = 1 + sqrt(19.6) t + (0.5) (9.8) t²                       

Y = 1 + sqrt(19.6) t + (0.5) (9.8) t²                             eq-1

for the second cannonball :

v_o = initial velocity = 0 m/s

t = time of travel

Y' = final position

using the equation

Y' = v_o t + (0.5) a t²

Y' = (0.5) (9.8) t²                 eq-2      

subtracting eq-2 from eq-1

Y - Y' = 1 + sqrt(19.6) t + (0.5) (9.8) t² - (0.5) (9.8) t²

Y - Y' = 1 + sqrt(19.6) t

hence the distance between the two balls is greater than 1

c)

for the first cannonball "

v_o = initial velocity = 5 m/s

v_f = final velocity = ?

h = height travelled

Using the equation

v_f² = v_o² + 2 a h

v_f² = 5² + 2 (9.8) h

v_f² = 25 + 2 (9.8) h                                       eq-1

for the second cannonball "

v_o = initial velocity = -5 m/s

v'_f = final velocity = ?

h = height travelled

Using the equation

v'_f² = v_o² + 2 a h

v'_f² = (-5)² + 2 (9.8) h

v'_f² = 25 + 2 (9.8) h                                       eq-2

subtracting eq2 from eq-1

v_f² - v'_f² = (25 + 2 (9.8) h) - (25 + 2 (9.8) h)

v_f² - v'_f² = 0

v_f = v_f'

hence the final velocities of two cannonball are same