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.

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.

The answer to your question is the last one

Horizontal change or the change of x