Question

standing on the edge of a cliff 32.5m high, you drop a ball. Later you throw a second ball downward with initial speed of 11m/s. Which ball has greater increase in speed when in reaches base of the cliff or do they travel at same speed. Verify your answer with a calculation.

Answer 1

Answer on Question #74126 Physics / Mechanics | Relativity

standing on the edge of a cliff $h = 32.5\mathrm{m}$ high, you drop a ball. Later you throw a second ball downward with initial speed of $u = 11\mathrm{m/s}$. Which ball has greater increase in speed when in reaches base of the cliff or do they travel at same speed. Verify your answer with a calculation.

Solution:

The displacement of the ball

$$h = \frac {v _ {f} ^ {2} - v _ {i} ^ {2}}{2 g}$$

So, final speed of the first ball

$$v _ {1} = \sqrt {2 g h} = \sqrt {2 \times 9 . 8 \times 3 2 . 5} = 2 5. 2 4 \frac {\mathrm {m}}{\mathrm {s}}$$

The change of it speed

$$\Delta v _ {1} = 2 5. 2 4 - 0 = 2 5. 3 4 \frac {\mathrm {m}}{\mathrm {s}}$$

The final speed of the second ball

$$v _ {2} = \sqrt {u ^ {2} + 2 g h} = 2 7. 5 3 \frac {\mathrm {m}}{\mathrm {s}}$$

The change of the second ball speed

$$\Delta v _ {2} = 2 7. 5 3 - 1 1 = 1 6. 5 3 \frac {\mathrm {m}}{\mathrm {s}}$$

Thus

$$\Delta v _ {1} > \Delta v _ {2}$$

Answer: $\Delta v_{1} > \Delta v_{2}$

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