a) how long did it take the ball to fall the first 10 cm? What was the average v

Question

a) how long did it take the ball to fall the first 10 cm? What was the average velocity for the period?

b) How fast were the balls traveling when they reached the 10 cm mark? What was their acceleration then?

c)About how fast was the light flashing (flashes per second)

The multiflash photograph in the accompanying figure shows two balls falling from rest. The vertical rulers are marked in centimeters. Use the equation s 49ot (the freefall equation for s in centimeters and tin seconds) to answer the following questions It takes sec for the balls to fall the first 10 cm. (Type an integer or a simplified fraction.) 10 cm (5 flashes) 40 cm (9 flashes 90 cm (13 flashes 60 cm (17 flashes)

Explanation / Answer

displacement, s=490*t^2

velocity, v=ds/dt

v=490*2*t

v=980*t m/sec

and

acceleration, a=dv/dt

a=980 m/sec^2

a)

s=490*t^2

10=490*t^2

===> t=0.143 sec

if t1=0.143 sec

==> s1=490*0.143^2

s1=10.02cm

if t2=0 sec,

====> s2=0

v_avg=s1-s2/t1-t2

=(10.02-0)/(0.143-0)

=70.07 cm/sec

b)

v=980*t

v=980*0.143

v=140.14 cm/sec

and

acceleration, a=980 cm/sec^2

c)

flash rate = total no flashes/time taken

=17/0.143

=118.88 flashes/sec

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