Answer:
[tex]30\; \rm m\cdot s^{-1}[/tex].
Explanation:
Under the assumptions, the acceleration of this ball would be constant during the descent. Thus, the SUVAT equations would apply.
Let [tex]x[/tex] denote the displacement of this stone during the fall. Let [tex]v_{0}[/tex] and [tex]v_{1}[/tex] denote the velocity of this ball at the beginning at the end of the fall, respectively. Let [tex]a[/tex] denote the acceleration of this ball. The SUVAT equation [tex]{v_{1}}^{2} - {v_{0}}^{2} = 2\, a\, x[/tex] would apply.
Rearrange the equation to find an expression for [tex]v_{1}[/tex], the velocity of this ball at the end of the fall:
[tex]\displaystyle {v_{1}}^{2} = 2\, a\, x + {v_{0}}^{2}[/tex].
If the final velocity of the ball is in the same direction as the displacement (downwards), then [tex]v_{1} > 0[/tex], such that [tex]\displaystyle v_{1} = \sqrt{2\, a\, x + {v_{0}}^{2}}[/tex].
The acceleration of this ball during the fall should be equal to the gravitational field strength: [tex]a = g = 10\; \rm m\cdot s^{-2}[/tex].
The displacement of the ball during the fall would be [tex]x = 45\; \rm m[/tex].
The question states that the ball was "dropped" from the given height. Thus, the initial velocity of the ball would be [tex]0\; \rm m\cdot s^{-1}[/tex]. (That is, [tex]v_{0} = 0\; \rm m \cdot s^{-1}[/tex].)
Substitute these values into the expression for [tex]v_{1}[/tex] and evaluate:
[tex]\begin{aligned}v_{1} &= \sqrt{2\, a\, x + {v_{0}}^{2}} \\ &= \sqrt{2 \times 10\; \rm m\cdot s^{-2} \times 45\; \rm m + 0\; \rm m \cdot s^{-1}} \\ &= 30\; \rm m\cdot s^{-1}\end{aligned}[/tex].
Therefore, the velocity of this ball right before landing would be [tex]v_{1} = 30\; \rm m\cdot s^{-1}[/tex].
