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 1.75 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.

Define

g = 9.8 m/s², acceleration due to gravity, positive downward.

Assume that wind resistance may be neglected.

Frog A:

u = 0.551 m/s, launch velocity, upward.

When the frog lands back on the pad, its vertical position is zero, and its vertical velocity will be 0.551 m/s downward.

If the time of flight is t, then

(0.551 m/s) * (t s) - 0.5 * (9.8 m/s²) * (t s) ² = 0

0.551t - 4.9t² = 0

t = 0, or t = 0.1124 s

t = 0 corresponds to launch, and t = 0.1124 s corresponds to landing.

Frog B:

Launch velocity is 1.75 m/s

When t = 0.1124 s, the position of the frog is

s = (1.75 m/s) (0.1124 s) - 0.5 * (9.8 m/s²) * (0.1124 s) ²

= 0.135 m

The velocity of frog B is

v = (1.75 m/s) - (9.8 m/s²) * (0.1124 s)

= 0.6485 m/s

Answer:

When frog A lands on the ground,

Frog B is 0.135 m above ground and its velocity is 0.649 m/s upward.