Two small frogs simultaneously leap straight up from a lily pad. Frog A leaps wi

Question

Two small frogs simultaneously leap straight up from a lily pad. Frog A leaps with an initial velocity of 0.551 m/s, while frog B leaps with an initial velocity of 1.23 m/s. When the first frog to return to the lily pad does so, what is the position and velocity of the other frog? Take upwards to be positive, and let the position of the lily pad be zero.

I first used 0=.551 + -9.8*t  to get time of A to be 0.05622s.
The height of A was then 0.1549m

Total time to go up and down for A is just 2x time which is .112448s to get back to the lilipad

I used v=1.23 + -9.8*0.112448 to get -0.12801 as the velocity of the Frog B at time when A hits the lilipad

Then, I used x = 1.23*0.112448 + 0.5*9.8*0.112448^2      to get 0.200269m  where Frog b is off the lilipad.

Both my answers for velocity and height is wrong for frog B...help please

Explanation / Answer

The time for frog A can be calculated from

vf = vo + at

-.551= .551 + (-9.8)(t)

t = .112 sec

Now we will use that time for frog B

vf = vo + at

vf = 1.23 + (-9.8)(.112)

vf = .128 m/s (Note its positive)

For the displacement

d = vot + .5at^2

d = (1.23)(.112) + (.5)(-9.8)(.112)^2

d = .076 m

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