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1.  Rob throws a baseball upwards at 14.2 m/s.  His friend, John, is sitting in a tree 4.5m above Rob.

a.  Calculate how long it will take to reach John.

b.  If John misses the ball as it moves upwards, how long will it take to reach John again.

Oct 3rd, 2015

1.Rob throws a baseball upwards at 14.2 m/s.  His friend, John, is sitting in a tree 4.5m above Rob.

a. Calculate how long it will take to reach John.

b. If John misses the ball as it moves upwards, how long will it take to reach John again?

Given u = 14.2 m/s, h = 4.5m, g = 9.8m/s2

a.  S= ut – 0.5gt2

4.5 = 14.2t -4.9t2

Rearranging and solving by quadratic equation

[img width="137" height="45" src="file:///C:/Users/NORMAX/AppData/Local/Temp/msohtmlclip1/01/clip_image001.gif" alt=" x=(-b+/-sqrt(b^2-4ac))/(2a). " v:shapes="Picture_x0020_1">

-4.9t2 + 14.2t – 4.5 = 0

t = 0.36 sec

b.  Max height, h = u2/ 2g

h = 14.22/ 2*9.8 = 10.29 m

distance to john position= 10.29-4.5 = 5.79

[img width="80" height="51" src="file:///C:/Users/NORMAX/AppData/Local/Temp/msohtmlclip1/01/clip_image002.png" alt="\ t =\ \sqrt {\frac{2d}{g}} " v:shapes="Picture_x0020_2">

t= sqrt( 2*5.79/9.8)

=1.08 sec.

Sep 21st, 2015

how did you find part b?

Sep 21st, 2015

In part b, first i determined max height which is 10.29m, at max hight the ball wil freely fall with an initial velocity of 0. since john is 4.5 m from ground, the ball will fall through a distance of 5.8 m to john taking 1.08 seconds

time to reach max height is sqrt (2h/g) = 1.45 sec, time to reach john = 0.36 sec, time to reach max height after  passing john = 1.45 -0.36= 1.09 sec, so total time to back to john= 1.08 + 1.09 = 2.17 sec

answer to part b= 2.17 sec

Regards

Sep 21st, 2015

In case you need more clarification or assistance in physics, kindly contact me. Thanks

Sep 21st, 2015

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