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?

Answer 1

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

Answer 2

Ok..I am not sure how to solve for t?

Answer 3

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