Solution :

Given :

Discharge ,$ Q=17.93 \mathrm{~m}^{3} / \mathrm{s}$

Depth before jump, $ d_{1}=0.8 \mathrm{~m}$

Taking width, $b=1 \mathrm{~m}$,

we get Discharge per unit width,

$ q=\frac{17.93}{1}=17.93$

Let

$ d_{2}=$ Depth after jump and

$h_{L}=$ Loss of energy.

we get

$\begin {aligned} d_{2} &=-\frac{d_{1}}{2}+\sqrt{\frac{d_{1}^{2}}{4}+\frac{2 q^{2}}{g d_{1}}} \\ &=-\frac{0.8}{2}+\sqrt{\frac{0.8^{2}}{4}+\frac{2 \times 17.93^{2}}{9.81 \times 0.8}} \\ &=-0.4+\sqrt{0.16+81.927} \\ &=-0.4+9.06 \\ &=8.66 \mathrm{~m} \end{aligned} $

for loss of energy,

$\begin {aligned} h_{L} &=\frac{\left(d_{2}-d_{1}\right)^{3}}{4 d_{1} d_{2}} \\ &=\frac{(8.66-0.8)^{3}}{4 \times 0.8 \times 8.66} \\ &=17.52 \mathrm{~m} \end{aligned}$