Question

By trial and error, a frog learns that it can leap a maximum horizontal distance of 1.5 m. If, in the course of an hour, the frog spends 33% of the time resting and 67% of the time performing identical jumps of that maximum length, in a straight line, what is the distance traveled by the frog?

Answer 1

Condition:

By trial and error, a frog learns that it can leap a maximum horizontal distance of $1.5\,\mathrm{m}$. If, in the course of an hour, the frog spends $33\%$ of the time resting and $67\%$ of the time performing identical jumps of that maximum length, in a straight line, what is the distance traveled by the frog?

Solution:

Let's find time per jump = T.

Frog's leap can be considered as a motion of projectile. For a maximum horizontal distance frog must leap at angle of $45{}^{\circ}$ to horizontal. So,

time per jump $T = \frac{2v_y}{g}$

horizontal distance $S = v_x T$ and $v_x = v_y$ for angle of $45{}^{\circ}$

We have $T = \frac{2v_y}{g} = \frac{2v_x}{g} = \frac{2S}{gT} \rightarrow T = \sqrt{\frac{2S}{g}} = \sqrt{2 * \frac{1.5}{9.8}} = 0.55\,\mathrm{s}$

Jumps per hour if jumping $100\%$ of the time $= \frac{60\,\mathrm{min}}{T}$

Total distance per hour if jumping $100\%$ of the time = (Jumps per hour) x (distance per jump) $= \left(\frac{60\,\mathrm{min}}{T}\right) \cdot (1.5\,\mathrm{m})$

Now multiply this quantity by $67\%$ (since the frog jumps only $67\%$ of the time)

So, total distance $= 0.67 \cdot \left(60\,\frac{\mathrm{min}}{T}\right) \cdot (1.5\,\mathrm{m}) = 0.67 \cdot \left(\frac{3600\,\mathrm{s}}{0.55\,\mathrm{s}}\right) \cdot 1.5\,\mathrm{m} = 6578\,\mathrm{m} \approx 6.6\,\mathrm{km}$

Answer: total distance $\approx 6.6\,\mathrm{km}$