posted by Kelly
An athlete performing a long jump leaves the ground at a 27.1∘ angle and lands 7.72m away
1.What was the takeoff speed?
2.If this speed were increased by just 5.0%, how much longer would the jump be?
u = speed * cos 27.1
7.72 = speed *cos 27.1 * t
where t is time in air
h = Hi + Vi t - (1/2) g t^2
0 = 0 + speed*sin 27.1 * t - 4.9 t^2
but we know t = 7.72/(speed*cos 27.1)
call speed s
0 = 0 + 7.72 tan 27.1 - 4.9*7.72^2/(s^2cos^227.1)
solve for s
do it again for s = 1.05 s