Question

Two small frogs simultaneously leap straight up from a lily pad. Frog A leaps with an initial velocity of 0.551 m / s, and frog B leaps with an initial velocity of 1.62 m / s. When frog A lands back on the lily pad, what are the position and velocity of frog B.

Answer 1

0.51800346 m/s, 0.12 m

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

`v=u+at\\\Rightarrow t=(v-u)/(a)\\\Rightarrow t=(-0.551-0.551)/(-9.81)\\\Rightarrow t=0.112334\ s`

Time taken by frog A is 0.112334 s

`v=u+at\\\Rightarrow v=1.62-9.81* 0.112334\\\Rightarrow v=0.51800346\ m/s`

`s=ut+(1)/(2)at^2\\\Rightarrow s=1.62* 0.112334+(1)/(2)* -9.81* 0.112334^2\\\Rightarrow s=0.12\ m`

The velocity of frog B is 0.51800346 m/s and the distance is 0.12 m