1. **State the problem:** We are given the velocity distribution for flow over a plate as $$u=34y - y^2$$ where $$u$$ is velocity in m/s and $$y$$ is the distance in meters above the plate. We need to find the shear stress at $$y=0.15$$ m. 2. **Recall the formula for shear stress:** Shear stress $$\tau$$ at a distance $$y$$ from the plate is given by $$\tau = \mu \frac{du}{dy}$$ where $$\mu$$ is the dynamic viscosity and $$\frac{du}{dy}$$ is the velocity gradient at $$y$$. 3. **Find the velocity gradient:** Differentiate $$u=34y - y^2$$ with respect to $$y$$: $$\frac{du}{dy} = 34 - 2y$$ 4. **Evaluate the velocity gradient at $$y=0.15$$:** $$\frac{du}{dy} \bigg|_{y=0.15} = 34 - 2(0.15) = 34 - 0.3 = 33.7$$ 5. **Convert dynamic viscosity from poise to Ns/m²:** Given $$8.5$$ poise and $$1$$ poise $$= 0.1$$ Ns/m², $$\mu = 8.5 \times 0.1 = 0.85\, \text{Ns/m}^2$$ 6. **Calculate shear stress:** $$\tau = \mu \frac{du}{dy} = 0.85 \times 33.7 = 28.645\, \text{Ns/m}^2$$ **Final answer:** The shear stress at $$y=0.15$$ m is $$\boxed{28.645\, \text{Ns/m}^2}$$.