Solution:
Given:
\[
y = 8t - 5t^2 \quad \text{and} \quad x = 6t
\]
To find the initial velocity, we calculate the horizontal and vertical components of velocity:
1. Horizontal velocity \( v_x = \frac{dx}{dt} = 6 \, \text{m/s} \)
2. Vertical velocity \( v_y = \frac{dy}{dt} = 8 - 10t \)
At \( t = 0 \) (initial velocity):
\[
v_y(0) = 8 \, \text{m/s}
\]
Thus, the magnitude of the initial velocity \( u \) is:
\[
u = \sqrt{v_x^2 + v_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s}
\]
The velocity of projection is \( 10 \, \text{m/s} \).
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