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.

Answer 1

Answer:

d = .076 m

Explanation:

The time for frog A can be calculated from  equation of motion

$v_f = v_o + at$

where v_f is final velocity, a is acceleration due to gravity

so from given data we have

$-0.551= 0.551 + (-9.8)(t)$

t = 0.112 sec

Now we will use that time for frog B

$v_f = v_o + at$

$v_f = 1.23 + (-9.8)(0.112)$

$v_f = 0.128 m/s$(Note its positive)

For the displacement

$s = v_o t + 0.5at^2$

$s = (1.23)(0.112) + (.5)(-9.8)(0.112)^2$

d = .076 m