(a) [tex]6.43\cdot 10^5 J[/tex]
The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):
[tex]E=K+U[/tex]
The initial kinetic energy is:
[tex]K=\frac{1}{2}mv^2[/tex]
where m = 58.0 kg is the mass of the projectile and [tex]v=140 m/s[/tex] is the initial speed. Substituting,
[tex]K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J[/tex]
The initial potential energy is given by
[tex]U=mgh[/tex]
where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,
[tex]U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J[/tex]
So, the initial mechanical energy is
[tex]E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J[/tex]
(b) [tex]-1.67 \cdot 10^5 J[/tex]
We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.
The kinetic energy is
[tex]K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J[/tex]
while the potential energy is
[tex]U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J[/tex]
So, the mechanical energy is
[tex]E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J[/tex]
And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:
[tex]W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J[/tex]
And the work is negative because air friction is opposite to the direction of motion of the projectile.
(c) 88.1 m/s
The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:
[tex]W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J[/tex]
When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:
E = K
The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:
[tex]E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J[/tex]
And this is only kinetic energy:
[tex]E=K=\frac{1}{2}mv^2[/tex]
So, we can solve to find the final speed:
[tex]v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s[/tex]
