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Projectile Motion

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iamdan08:
I was looking at some questions from the "Jacaranda Physics 2" textbook (i use heinmenn) and i came across this question:

A gymnast wants to jump a distance of 2.5m, leaving the ground at an angle of 28 degrees. With what speed must the gymnist take off?

Any help would be great   ;D

ed_saifa:
i use the same book! the questions are really ambiguous and this one is no different.

iamdan08:
lol...i agree. I have tried approaching the questions a few different ways. Can't seem to get it though!

unknown id:
To answer this question, its better to use only the first half of the parabolic path taken by the gymnast. Hence, it's best to start off by listing all your known values:

uh = ucos(28) m/s ---> can be derived from a vector diagram
uv = usin(28) m/s

ah = 0 m/s^2
av = -10 m/s^2

vh = ucos(28) m/s ---> stays the same
vv = 0 m/s

xh = 1.25 m

Note: xh, vh and vv are the values taken at the end of half of the parabolic path.

Now, we can use the constant acceleration formulae to calculate the time taken by the gymnast to get to the maximum turning point of her path. The following calculation makes use of horizontal projectile values:







Now that we've got a value for time, we can substitute this value into another formula. The following calculation makes use of vertical projectile values:

iamdan08:
Awsome! Thats the right answer.  :D  How do we know that its the horizontal speed that needs to be found and not the speed at the angle of 28 degrees? Do we just assume?

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