We make a graphic of this problem to define the angle.
The angle we can calculate through triangle relation, that is,
With this function we should only calculate the derivate in function of c
That is the rate of change of 
.
b) At this point we need only make a substitution of 0 for c in the equation previously found.
Hence we have finally the rate of change when c=0.
<span>The initial speed, u of plane in terms of velocity of sound which may be taken as U
u=142/331=0.429*U
It crosses the sound barrier after says t seconds then we have 331-142=23.1*t or t is given 8.18 s exactly t=9/11s.
After 18 seconds the plane will traveling with velocity V
V=142+18*23.1=557.8 m/s==1.685*U</span>
Time = distance divided by speed
4,900 divide 1.5 = 3266.67
SI unit for mas is Kg and the SI unit of weight N for newton
m = mass of the partner which the cheerleader lifts = 59.6 kg
h = height to which the partner is lifted by the cheerleader = 0.749 m
g = acceleration due to gravity = 9.8 m/s²
work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.
W = work done by the cheerleader in lifting the partner
PE = potential energy gained
so W = PE
potential energy is given as
PE = mgh
hence
W = mgh
inserting the values in the above formula
W = 59.6 x 9.8 x 0.749
W = 437.5 J
this is the work done in lifting the partner once.
the cheerleader does this 30 times , hence the total work done is given as
W' = 30 W
W' = 30 x 437.5
W' = 13125 J
