A shotputter throws the shot with an initial speed of 16.0 m/s at a 36.0 angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.00 m above the ground
3 answers
Hmm, I hate physic's, it messes with your head.
Solve this equation for t.
Y = 2.00 + 16sin36*t - (g/2)t^2
and take the positive root for t.
g = 9.81 m/s^2, of course
Then multipy t by the horizontal velocity component, 16.0 cos 36 = 12.94 m/s
Y = 2.00 + 16sin36*t - (g/2)t^2
and take the positive root for t.
g = 9.81 m/s^2, of course
Then multipy t by the horizontal velocity component, 16.0 cos 36 = 12.94 m/s
36.1 is the answer i got
its wrong though
what could i have done wrong?
its wrong though
what could i have done wrong?