Hmm, I may be overlooking something here, unless the mass is irrelevant, but I drew Y0=0 V0=8.1m/s VF=0m/s( because the jumper reached Yf stops and comes back down toward earth) and a=9.8m/s^2 so you're looking for Y final position
I used Vf^2=V0^2+2a(Yf-Y0) since vf=0 and Y0=0 rearrange to solve for Yf
-Vo^2/2a=yf
-8.1m/s)^2/2(9.8m/s^2)= Yf
65.61m^2/s^2/19.6m/s^2
Yf=3.35m .....
The only reason I could think of that the mass is here is because F=ma
A 65.5 kg high jumper leaves the ground with a vertical velocity of 8.1 m/s.
How high can he jump? The acceleration of gravity is 9.8 m/s2 .
2 answers
The mass is irrelevant
(1/2) m v^2 = m g h
note mass cancels
h = (1/2) v^2 / g = 3.347 = 3.35 meters
correct
(1/2) m v^2 = m g h
note mass cancels
h = (1/2) v^2 / g = 3.347 = 3.35 meters
correct