*** For higher ratings/Best Answer: please explain how you find your answer (sho
ID: 1276600 • Letter: #
Question
*** For higher ratings/Best Answer: please explain how you find your answer (show your work, draw a picture). Let me know what equations you used and what the different variables stand for/were in the problem.I really need to understand how to do this problem, and other ones like it. Thanks.
Here's the question:
In the Olympic shotput event, an athlete throws the shot with an initial speed of 12.0 m/s at a 40.0degree angle from the horizontal. The shot leaves her hand at a height of 1.80 m above the ground. How far does the shot travel? Repeat the calculation of part (a) for angles 42.5degree, 45.0degree, and 47.5degree.Explanation / Answer
a) v = 12 m/s, theta = 40.0o, h = 1.8 m
vertical displacement = -h = v sin theta * t - gt2/2
-1.8 = 7.7135 t - 4.9t2
4.9t2 - 7.7135t - 1.8 = 0
t = 1.78 s
horizontal displacement = v cos? * t
= 12 * cos 40 * 1.78
= 16.36 m
----------------------------------------------------
b) theta = 42.50
horizontal displacement = v cos? * t
= 12 * cos 42.5 * 1.78
= 15.75 m
------------------------------------
theta = 450
horizontal displacement = v cos? * t
= 12 * cos 45 * 1.78
= 15.10 m
-------------------------------------
theta = 47.50
horizontal displacement = v cos? * t
= 12 * cos 47.5 * 1.78
= 14.43 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.