Physics problem - part 5

A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.

From the equations of motion we have

v = u - gt
v = final velocity; u = initial velocity; g = acceleration due to gravity; t = time of flight.

Vertical height = ut - (1/2) g t^2

The trick to this and other problems of a similar kind is to realise that at the top of the trajectory, the vertical velocity is momentarily zero. We can use this fact to work out how long it takes to reach maximum height. So

0 = u - gt
gt = u
t = u/g

This is the time it takes to reach its max height. We now plug this value of time into the equation for its height....
Height = ut - (1/2) g t^2

Height = u[u/g] - (1/2) g [u^2/g^2]
Height = u^2/g - (1/2) u^2/g = (1/2) u^2/g

2.62 = u^2/2g
u^2 = 2g * 2.62
u^2 = 2 * 9.8 * 2.62 = 51.352

u = sqrt(51.352) = 7.17 m/s

Using this formula Vf²=Vi² + 2ad
Therefore: Vi²= Vf² - 2ad
Vi = √(Vf² - 2ad)
We have Vf=0m/s ( It's a final speed and when it jump to highest point, final speed is 0m/s)
We have a= -9.8 m/s²( It's a gravitational force and it's down force, so it must be negative 9.8 m/s²)
We have d= 2.62m ( It's a distance that it jump, so it must be the height 2.62 m)

Vi= √ (0 - 2(-9.8)(2.62)) = 7.17m/s ( 3 s.f) ( Use 3s.f because that what they gave us)

vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)

0 m2/s2 = vi2 - 51.35 m2/s2

51.35 m2/s2 = vi2

vi = 7.17 m/s

g=-10m/s
v=0
h=2.62m
v²=u²+2gh
0=u²-2x10x2.62
u²=20x2.62
u²=52.4
u=7.23m/s