4. A particularly athletic frog can push herself off of the ground with enough v

Question

4. A particularly athletic frog can push herself off of the ground with enough velocity to jump 1m high. Note that the initial velocity of the frog, once you find it, is a property of the frog: no matter what we do with the frog, the initial velocity (which determines how high she can jump) is the same. (a) With what velocity does the frog leave the ground? (b) How long will the frog be in the air before she lands?

Suppose we now put the frog in an elevator, accelerating upward at = 2 m/s 2 . There is a tasty spider at a height h = 85 cm above the floor of the elevator. This is the same frog, so it is capable of jumping with an initial velocity equal to the value you found in the previous problem. (a) If the frog jumps as high as she can, will she catch the spider? How far above the elevator floor will the frog make it? (b) How long will the frog be in the air? Hint: Think very carefully about your coordinate system, and all of the consequences of the accelerating elevator. You may need the quadratic formula for this problem. If you are still stuck, draw position vs. time graphs for the frog and the spider.

Explanation / Answer

H = maximum height (for a vertical jump) = (u)²/(2g)

1 = [u]²/[2(9.81)]

u = [19.62] = 4.43 m/s (vertically upwards)

time of flight = 2[u] / g = 0.903047 s

In the elevator when the frog jumps relatively to the spider needs to catch the spider within its capacity of initial velocity of 4.43 m/s.

[please note: if the the velocities of objects A(frog) and B(spider) are v and v with respect to a stationary observer (neutral observer) on the ground surface, then the velocity of A relative to B is v - v. this is as if B(spider) were at rest and only A were moving.]

initial v - v = (4.43 + v) - v = +4.43 [initially before the jump both are moving upwards at the same v]

relative acceleration = a - a = -9.81 - (2) = -11.81 m/s²

relatively the spider is at rest, 0.85 m above A, and A(frog) is moving with an initial velocity 4.43 m/s upwards and acceleration -11.81 m/s² (downwards) towards it

relatively applying s = ut + ½at²

0.85 = ut + ½at²

0.85 = 4.43t + ½(-11.81)(t)²

t² - 0.75t + 0.144 = 0

solving there are no real solutions to this(only imaginary), hence the frog can’t catch spider!

again relatively applying v² = u² + 2as(noting that relatively the frog will have 0 velocity at the highest point w.r.t the elevator floor)

0² = 4.43² - 2(11.81)s

s = 0.831 m

(considering the elevator floor is at rest and only frog moving)

also relatively when the frog returns it will have the same speed 4.43 m/s only directed downwards

relatively applying v = u + at

-4.43 = 4.43 + (-11.81)t

t = 8.86 / 11.81 = 0.750 s

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