How Can a Grasshopper Jump 1 Meter Starting from Rest?

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To determine how a grasshopper can jump 1 meter from rest at a 45-degree angle, the initial speed required is calculated to be 3.13 m/s, and the maximum height reached is 0.250 meters. The discussion emphasizes the use of kinematic equations and the relationship between the jump's range, angle, and initial speed. A participant expresses confusion about deriving the initial speed and height from the given parameters, particularly struggling with the concept of starting from rest. The conversation highlights the need for clear step-by-step guidance in applying the relevant formulas to solve the problem effectively. Understanding these principles is essential for progressing in the topic of projectile motion.
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Homework Statement



A grasshopper jumps 1.00 m from rest, with an initial velocity at a 45.0 degree angle with respect to the horizontal. Find a. the initial speed of the grasshopper (the answer is 3.13 m/s) and b. the maximum height reached ( the answer is .250 meters).

Homework Equations



V naught of x= v naught (hyp. of v. vectors) times cos (projection angle)
It should all be derived from the basic mechanic equations;
V= vnaught times time + a times time
Delta x= vnaught times time + .5 a times time squared
v squared= v naught squared + 2a delta x
I suspect they want me to use the cosine and sine versions but I need the steps so I can truly understand what I'm looking for!

The Attempt at a Solution


I keep getting zero I don't understand

Can somebody please give me all the steps for this problem and what formulas to use if I don't have them. I can crunch the numbers but for some reason I have a mental block on this problem. I don't even get how if it starts at rest, it can have any other value but zero.
 
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Who said:

Homework Statement



A grasshopper jumps 1.00 m from rest, with an initial velocity at a 45.0 degree angle with respect to the horizontal. Find a. the initial speed of the grasshopper (the answer is 3.13 m/s) and b. the maximum height reached ( the answer is .250 meters).

Homework Equations



V naught of x= v naught (hyp. of v. vectors) times cos (projection angle)
It should all be derived from the basic mechanic equations;
V= vnaught times time + a times time
Delta x= vnaught times time + .5 a times time squared
v squared= v naught squared + 2a delta x
I suspect they want me to use the cosine and sine versions but I need the steps so I can truly understand what I'm looking for!

The Attempt at a Solution


I keep getting zero I don't understand

Can somebody please give me all the steps for this problem and what formulas to use if I don't have them. I can crunch the numbers but for some reason I have a mental block on this problem. I don't even get how if it starts at rest, it can have any other value but zero.
Write out the expression for v as a function of range R, and the angle. You have enough information to calculate it. see: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra13

AM
 
That really doesn't help me. I can't find the range without at least the V naught of the y component and the range of both the x and y component velocity vectors. So, at halfway the velocity component in the x direction is 0 and the vertical component of that is the velocity of Y at .5 meters which still leads me back to the formula where I am without V naught of R. I really just need the steps because then I can work backwards, without them if I try to move on in the chapter without this knowledge it does me no good.
 
Who said:
That really doesn't help me. I can't find the range without at least the V naught of the y component and the range of both the x and y component velocity vectors. So, at halfway the velocity component in the x direction is 0 and the vertical component of that is the velocity of Y at .5 meters which still leads me back to the formula where I am without V naught of R. I really just need the steps because then I can work backwards, without them if I try to move on in the chapter without this knowledge it does me no good.
It certainly does help you. Do you think you are expected to work out the relationship between range, angle and speed every time you have such a problem? You have the range, R. You have the angle. You know g. All you have to do is plug these values into the equation!

v_o^2 = Rg/sin(2\theta)

AM
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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