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

Answer:

0.51800346 m/s, 0.12 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

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

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{-0.551-0.551}{-9.81}\\\Rightarrow t=0.112334\ s$

Time taken by frog A is 0.112334 s

$v=u+at\\\Rightarrow v=1.62-9.81\times 0.112334\\\Rightarrow v=0.51800346\ m/s$

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=1.62\times 0.112334+\dfrac{1}{2}\times -9.81\times 0.112334^2\\\Rightarrow s=0.12\ m$

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