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madreJ [45]
3 years ago
12

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

ntal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground. (See Nature, Vol. 424, July 31, 2003, p. 509.) (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?
Physics
1 answer:
Jobisdone [24]3 years ago
7 0

(a) 4.0 m/s

We can solve this part just by analyzing the vertical motion of the froghopper.

The initial vertical velocity of the froghopper as it jumps from the ground is given by

u_y = u_0 sin \theta (1)

where

u_0 is the takeoff speed

\theta=58.0^{\circ} is the angle of takeoff

The maximum height reached by the froghopper is

h = 58.7 cm = 0.587 m

We know that at the point of maximum height, the vertical velocity is zero:

v_y = 0

Since the vertical motion is an accelerated motion with constant (de)celeration g=-9.8 m/s^2, we can use the following SUVAT equation:

v_y^2 - u_y^2 = 2gh

Solving for u_y,

u_y = \sqrt{v_y^2-2gh}=\sqrt{-2(-9.8)(0.587)}=3.4 m/s

And using eq.(1), we can now find the initial takeoff  speed:

u_0 = \frac{u_y}{sin \theta}=\frac{3.4}{sin 58.0^{\circ}}=4.0 m/s

(b) 1.47 m

For this part, we have to analyze the horizontal motion of the froghopper.

The horizontal velocity of the froghopper is

u_x = u_0 cos \theta = (4.0) cos 58.0^{\circ} =2.1 m/s

And this horizontal velocity is constant during the entire motion.

We now have to calculate the time the froghopper takes to reach the ground: this is equal to twice the time it takes to reach the maximum height.

The time needed to reach the maximum height can be found through the equation

v_y = u_y + gt

Solving for t,

t=-\frac{u_y}{g}=-\frac{3.4}{9.8}=0.35 s

So the time the froghopper takes to reach the ground is

T=2t=2(0.35)=0.70 s

And since the horizontal motion is a uniform motion, we can now find the horizontal distance covered:

d=u_x T = (2.1)(0.70)=1.47 m

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Answer:50.39 m/s,

40.46 m

Explanation:

Given

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We know that Range =\frac{u^2sin2\theta }{g}

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H=40.46 m

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Answer:

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Answer:

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

Explanation:

Given that the child accelerates uniformly and that both initial (v_{o}) and final speeds (v_{f}), measured in meters per second, and acceleration (a), measured in meters per square second, are known, we proceed to use the following kinematic equation to determine the time taken to travel from the top of the hill to the bottom (t), measured in seconds, is:

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If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 17.5\,\frac{m}{s} and a = 2.94\,\frac{m}{s^{2}}, then the time taken is:

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