Question

a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertical cliff, what is the the hight of the cliff

Answer 1

Answer:

Approximately $281.25\; \rm m$. (Assuming that the drag on this ball is negligible, and that $g = 10\; \rm m \cdot s^{-2}$.)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

• Horizontal: no acceleration, velocity is constant (at $v(\text{horizontal})$ is constant throughout the descent.)
• Vertical: constant downward acceleration at $g = 10\; \rm m \cdot s^{-2}$, starting at $0\; \rm m \cdot s^{-1}$.

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: $x(\text{horizontal}) = 30\; \rm m$. Combine these two quantities to find the duration of this descent:

$\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}$.

In other words, the ball in this question start at a vertical velocity of $u = 0\; \rm m \cdot s^{-1}$, accelerated downwards at $g = 10\; \rm m \cdot s^{-2}$, and reached the ground after $t = 7.5\; \rm s$.

Apply the SUVAT equation $\displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t$ to find the vertical displacement of this ball.

$\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}$.

In other words, the ball is $281.25\; \rm m$ below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be $281.25\; \rm m\!$.