The best leaper in the animal kingdom is the puma, which can jump to a height of

varaderiyw

varaderiyw

Answered question

2021-11-16

The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45. With what speed must the animal leave the ground to reach that height?

Answer & Explanation

Pulad1971

Pulad1971

Beginner2021-11-17Added 22 answers

This problem aims to find the velocity of the puma at it leaves the ground, where it will be able to reach 3.7 m given an angle of 45
First, let us analyze the problem. Since it is jumping at an angle of 45, it's motion will follow a curved path. We know that at maximum height on acurved path, Vy=0
Thus, with the given from the problem, we can use the kinematic equation:
Vy2=Voy2+2ayy
where:
Vy= Final velocity
Voy2= Initial velocity
y= Displacement
ay=gravity=(9.8ms2)
Voy=Vy22ayy
Substituting in the given values, we can now calculate:
Voy=(9.8ms2)(3.7m0)2
Voy=8.52ms
Finally, to calculate the velocity at an angle, for vertical components we use the formula:
Voy=Vosinθ
Vo=8.52mssin45
Vo=12.05ms
RizerMix

RizerMix

Expert2023-05-27Added 656 answers

Given:
Maximum height (h) = 3.7 m
Launch angle (θ) = 45 degrees
The vertical component of the initial velocity can be determined using the equation:
vy=v·sin(θ)
where v is the magnitude of the initial velocity. We can rearrange this equation to solve for v:
v=vysin(θ)
Substituting the given values, we have:
v=3.7sin(45)
To evaluate this expression, we can use the fact that sin(45)=22:
v=3.722
Next, we simplify the expression by multiplying the numerator and denominator by 222:
v=3.7·222
Simplifying further:
v=3.7·22
Therefore, the puma must leave the ground with a velocity of approximately v=5.24m/s to reach a height of 3.7 m.
Vasquez

Vasquez

Expert2023-05-27Added 669 answers

The vertical motion of the puma can be described by the equation:
y=v0sin(θ)·t12gt2
where y is the vertical displacement (height) and g is the acceleration due to gravity. We know that the puma reaches a height of 3.7m, so we can write:
3.7=v0sin(45)·t12gt2
The horizontal motion of the puma can be described by the equation:
x=v0cos(θ)·t
where x is the horizontal displacement. Since the puma leaves the ground and lands at the same horizontal position, we can assume the horizontal displacement is zero:
x=0=v0cos(45)·t
Solving the second equation for t, we get:
t=0v0cos(45)=0
Substituting this value of t into the first equation, we can solve for v0:
3.7=v0sin(45)·012g·02=0
Therefore, we have a contradiction. The equation cannot be solved for a nonzero value of v0 since the time of flight is zero.
In conclusion, there is no initial velocity at which the puma can leave the ground and reach a height of 3.7m when launched at an angle of 45.
Don Sumner

Don Sumner

Skilled2023-05-27Added 184 answers

Step 1:
To solve this problem, we can use the principles of projectile motion. We need to find the initial velocity v at which the puma must leave the ground to reach a height of 3.7 m when launched at an angle of 45 degrees.
The height reached by a projectile can be determined using the equation:
h=v2sin2(θ)2g
Where:
- h is the maximum height reached,
- v is the initial velocity,
- θ is the launch angle (in radians),
- g is the acceleration due to gravity.
In this case, h=3.7m and θ=45=π4radians. The acceleration due to gravity is g=9.8m/s2.
Step 2:
Substituting these values into the equation, we have:
3.7=v2sin2(π4)2·9.8
Simplifying further, we get:
3.7=v2·(22)22·9.8
Simplifying and solving for v2, we have:
v2=3.7×2·9.8(22)2
Solving this equation, we find:
v2=72.728
Taking the square root of both sides, we get:
v=72.7288.526m/s
Therefore, the puma must leave the ground with a speed of approximately 8.526 m/s to reach a height of 3.7 m when launched at an angle of 45 degrees.

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