a high jumper approaches the takeoff point at a speed of 7.20 m/s. Assuming that only this speed determines the height to which he can rise, find the maximum height at which the jumper can clear the bar.
3 answers
im sorry but i cant tell you the answer ill be gladly to check though
KE runner=PE change of runner
1/2 m v^2=mgh solve for h.
This is a bit misleading, the runners center of gravity starts near his waist, about 1.2 meters above the ground. So, I would add that to h. Your teacher probably wouldn't
1/2 m v^2=mgh solve for h.
This is a bit misleading, the runners center of gravity starts near his waist, about 1.2 meters above the ground. So, I would add that to h. Your teacher probably wouldn't
so would the m's cancel out and leave 1/2 v^2= gh
1/2 (7.20)^2=9.8h
25.92=9.8h
h=2.64
1/2 (7.20)^2=9.8h
25.92=9.8h
h=2.64
3 answers