The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can accelerate at 4.00 km/s2 over a distance of 2.00 mm as it straightens its specially adapted“jumping legs.” Assume the acceleration is constant. (a) Find the upward velocity with which the insect takes off. (b) In what time interval does it reach this velocity? (c) How high would the insect jump if air resistance were negligible? The actual height it reaches is about 70 cm, so air resistance must be a noticeable force on the leaping froghopper.
I don't understand part C please help
I solved part A and B
I don't know how to start part C
x = v^2/2g
You found v in part b
I don’t understand the No.c
To understand part C of the question, let's break it down step by step.
First, we need to find the time it takes for the froghopper to reach its maximum upward velocity. We can do this using the formula:
v = u + at
where:
v is the final velocity (which in this case is the maximum upward velocity),
u is the initial velocity (which is 0 in this case, as the froghopper starts from rest),
a is the acceleration, and
t is the time.
We know the acceleration (a) is 4.00 km/s^2 and the distance (s) it covers during acceleration is 2.00 mm. We need to convert the distance to meters for consistency, so 2.00 mm is equal to 0.002 m.
Using the formula of motion:
s = ut + (1/2)at^2
where:
s is the distance covered, we can solve for time (t) to find the time it takes to cover the distance during acceleration.
Plugging the values into the formula, we have:
0.002 m = (1/2) × 4.00 km/s^2 × t^2
Now we can solve for t:
0.002 m = 2.00 m/s^2 × t^2
t^2 = 0.001
t ≈ 0.0316 s
Therefore, it takes approximately 0.0316 seconds for the froghopper to reach its maximum upward velocity.
Now, to calculate the maximum height the froghopper would jump if air resistance were negligible, we can use the equation:
v^2 = u^2 + 2as
Here, v is the maximum upward velocity (which we will calculate in the next step), u is again the initial velocity (0 m/s), a is the acceleration (4.00 km/s^2), and s is the total distance covered.
We need to calculate v, the maximum upward velocity, which is the velocity at the end of the acceleration phase and just before the froghopper starts descending.
Using the formula:
v = u + at
v = 0 + (4.00 km/s^2 × 0.0316 s)
v = 0.1264 km/s (or 126.4 m/s)
Now, substituting these values back into the equation for maximum height, we have:
(126.4 m/s)^2 = (0 m/s)^2 + 2(4.00 km/s^2)s
Simplifying the equation, we get:
15958.496 m^2/s^2 = 8000 m/s^2 × s
s ≈ 1.9951 m
Therefore, if air resistance were negligible, the froghopper would jump to a height of approximately 1.9951 meters.
However, the actual height the froghopper reaches is only about 70 cm or 0.7 meters. This indicates that air resistance has a significant impact on the trajectory and height of the froghopper's jump.