time in air:
distance=v*cos30*t
t= 8.9/v*.866
take off speed:
v*sin30 is vertical velocity (vi')
hf=hi+vi'*t-4.8t^2 solve for v, take off speed, using time in air t as above.
0=0=v*sin30*(8.9/v*.866)-4.8(8.9/v*.866)^2
or v^2*.5*8.9/.866=4.8*(8.9/.866)^2
v^2=4.8*8.9/.866 check my math, take square root
an athlete executing a long jump leaving the ground at an angle of 30° and travels 8.9m .what was the take off speed?
1 answer