Asked by Jim
A longjumper running at 12 m/s and jumps at 20 degrees. No wind or resistance but gravity is normal -9.81 m/s2. How far will person jump?
Answers
Answered by
bobpursley
Figure his time in air first.
Because time in air is determined by gravity, consider the vertical first.
Vertical: hf=ho+Vi*t-4.9t^2
0=0+12Sin20*t-4.9t^2
solve for time in air, t.
Now, the problem, how far does he go.
distance=horizontalvelocity*time
= 12cos20*timeinair
Because time in air is determined by gravity, consider the vertical first.
Vertical: hf=ho+Vi*t-4.9t^2
0=0+12Sin20*t-4.9t^2
solve for time in air, t.
Now, the problem, how far does he go.
distance=horizontalvelocity*time
= 12cos20*timeinair
Answered by
Jim
Ok..I am not sure how to solve for t?
Answered by
bobpursley
You may need to drop physics, the language of physics is math, and this is a pretty basic algebra I skill.
0=0+12sin20*t-4.9t^2
0=t(12sin20-4.9t)
so either t is zero, or t is 12sin20/4.9
0=0+12sin20*t-4.9t^2
0=t(12sin20-4.9t)
so either t is zero, or t is 12sin20/4.9
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