15. A hiker inspects a tree frog sitting on a small stick in his hand. Suddenly

Question

15. A hiker inspects a tree frog sitting on a small stick in his hand. Suddenly startled, the hiker drops the stick from rest at a height of 1.85 m above the ground and, at the same instant, the frog leaps vertically upward, pushing the stick down so that it hits the ground 0.450 s later. Find the height of the frog at the instant the stick hits the ground if the frog and the sticlk have masses of 7.25 g and 4.50 g, respectively. (Hint: Find the center-of-mass height at 1 0.450 s for the frog-stick system and then use the definition of center of mass to solve for the frog's height.)

Explanation / Answer

Initial Position of Center of mass = 1.85 m above ground

As g acts on the whole system, final position of CM at 0.45s = 1.8- 0.5*g*t*t =1m above ground

Position of Center of Mass of system= (Mass of Rod* Position of Rod + Mass of Frog*Position of Frog)/Total Mass

Therefore, position of frog = 1.62 m above ground

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