Answer:

\(u=160\).

Let the angle of projection be \(\theta\). Since

\(\sin\theta=\frac45,\)

we have

\(\cos\theta=\frac35.\)

The horizontal component of velocity remains constant throughout the motion:

\(v_x=u\cos\theta=\frac35u.\)

The vertical component at time \(t\) is

\(v_y=u\sin\theta-gt=\frac45u-gt.\)

At \(t=8\), the velocity components are

\(\left(\frac35u,\frac45u-8g\right).\)

At \(t=32\), the velocity components are

\(\left(\frac35u,\frac45u-32g\right).\)

The two directions of motion are perpendicular, so the scalar product of these two velocity vectors is zero:

\(\left(\frac35u\right)^2+\left(\frac45u-8g\right)\left(\frac45u-32g\right)=0.\)

Expanding,

\(\frac{9}{25}u^2+\frac{16}{25}u^2-\frac{160}{5}gu+256g^2=0.\)

So

\(u^2-32gu+256g^2=0.\)

Using \(g=10\),

\(u^2-320u+25600=0.\)

This is

\((u-160)^2=0.\)

Therefore

\(u=160.\)