physics problem?

Question

A certain frog's hind legs produce a force of 2.5 times its weight through a vertical distance of 9 cm, at which point the frog becomes airborne. A) What is the frog's speed at takeoff? B)What height about the ground does the frog reach?

the answers are 1.63 m/s and 13.5 cm

Please help me and show me how to do this
thanks!!

Answer 1

F=m*a

Looking at a FBD of the frog during launch

m*a=2.5*m*g-m*g
a=1.5*g

The easiest way to solve this is conservation of energy

.5*m*v^2=.09*1.5*m*g
v=sqrt(3*.09*9.81)
v=1.63 m/s

once airborne use

v(t)=vi-g*t
and
y(t)=vi*t-.5*g*t^2

apogee occurs when v(t)=0
t=vi/g
so the height of apogee is
y(apogee)=vi^2/g-.5*vi^2/g
or
=.5*vi^2/g
=13.5 cm

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