An athlete executing a long jump leaves the ground at a 27.7∘ angle and travels 7.60m.If this speed were increased by just 5.0%, how much longer would the jump be?

Answer 1

Vi = v sin 27.7
0 = Vi - 9.8 t

t = v sin 27.7 /9.8 that is time to the top
total time in air = 2 t
= 2 v sin 27.7/9.8

u = v cos 27.7
d = (v cos 27.7)(2 v sin 27.7 /9.8)
d = (2 v^2/9.8) cos 27.7 sin 27.7
but 2 cosA sinA = sin 2A
d = (v^2/9.8) sin (55.4)
either do d for v = 7.6 and v=1.05(7.6)
or take derivative
d d/dv = (2v sin 55.4)/9.8
d d = ( 2 v sin 55.4 /9.8) dv