Description
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.
Explanation & Answer
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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?
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
for free fall, u=0
[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.
Please let me know if you need any clarification. I'm always happy to answer your questions.
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