# Finding Velocity, time, distance

• Mar 21st 2010, 01:55 PM
Tiger
Finding Velocity, time, distance
An Olympic long jumper jumps 8.04m, reaching a height of 0.80 m halfway through his jump. What was his velocity as he left the ground? Also, how far would he be able to jump on the moon (g=1.63 N/kg) and how much time would he spend off the lunar surface?
• Mar 22nd 2010, 09:59 AM
Haytham
you need to separate the vertical component from the horizontal

you use the vertical to find out the time he was in the air

its easier to calculate the time he took to drop from the highest point (0.8) till he got to the ground and then multiply by 2
$\displaystyle 0=0.8-\frac{1}{2} g t^{2}$
t=0.4s
total time 0.8s

the vertical speed is simply
$\displaystyle 0.8=0.4v_y-\frac{1}{2} g 0.4^{2}$
$\displaystyle v_y=4m/s$
you can also use energy conservation to find it

next you use the horizontal data to find out the horizontal speed
$\displaystyle v_x=\frac{d}{t}$
$\displaystyle v_x=\frac{8.04}{0.8}=10.05m/s$

so $\displaystyle v=10.05 \vec{e_x}+4 \vec{e_y} (m/s)$

its similar for the moon, only changes g value