Physics Question?

Question

An athlete executing a long jump leaves the ground at a 30 degree angle and travels 7.80 m. What was the takeoff speed? If this speed were increased by just 5.0 percent, how much longer would the jump be? Please show/explain all work clearly.

Answer 1

First, the flight time is twice the time to apogee
vy(t)=sin(30)*v-g*t
apogee is when
vy(t)=0
so
t=sin(30)*v/g
and the flight time is
v/g

since we know that
x(t)=v*cos(30)*t
and the range is
7.8=v^2*cos(30)*sin(30)/g
and 2*cos(th)*sin(th)=sin(2*th)
v=sqrt(7.8*2*g/sin(60))
=13.3 m/s

If the new v is 1.05*v
then the new range is
D=1.05^2*v^2*cos(30)*sin(30)/g

D/7.80=1.05^2

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