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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 1.62 m/s. when frog a lands back on the lily pad, what is the position and velocity of frog b? take upwards to be positive, and let the position of the lily pad be zero.

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Kalahira
We need to find the time t required for frog A to reach maximum height. We can use this equation:
   
v = v_0 + at
t = (v - v_0) / a
t = (0 - 0.551 m/s) / (-9.80 m/s^2)
t = 0.05622 s
       
The total time for frog A to go up and come back down to the lily pad is 2t, which is 0.11245 s
       
We can use this time to find the position of frog B.
y = v0 t + 0.5 a t^2
y = (1.62 m/s)(0.11245 s) - (0.5)(9.80 m/s^2)(0.11245 s)^2
y = 0.12 m
       
We can use this time to find the velocity of frog B.
v = v0 + at v = 1.62 m/s - (9.80 m/s^2)(0.11245 s)
v = 0.518 m/s
       
The position of frog B is 0.12 meters above the lily pad.
The velocity is 0.518 m/s.

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