Answers: 2
. (Assuming that the drag on this ball is negligible, and that 
.)
is constant throughout the descent.)Vertical: constant downward acceleration at 
.
. Combine these two quantities to find the duration of this descent:
.
, accelerated downwards at 
.
to find the vertical displacement of this ball.![\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}](/tpl/images/2315/9405/72c91.png)
.
.
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