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&nbsp;45∘. With what speed must the animal leave the ground to reach that height?

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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&amp;nbsp;45∘. With what speed must the animal leave the ground to reach that height?

Pulad1971 · Accepted Answer

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&#039;s motion will follow a curved path. We know that at maximum height on acurved path, Vy=0Thus, with the given from the problem, we can use the kinematic equation:Vy2=Voy2+2ay△ywhere:Vy= Final velocityVoy2= Initial velocity△y= Displacementay=gravity=(−9.8ms2)Voy=Vy2−2ay△ySubstituting in the given values, we can now calculate:Voy=−(−9.8ms2)(3.7m−0)⋅2Voy=8.52msFinally, to calculate the velocity at an angle, for vertical components we use the formula:Voy=Vosin⁡θVo=8.52mssin⁡45∘Vo=12.05ms

RizerMix · Accepted Answer

Given:Maximum height (h) = 3.7 mLaunch angle (θ) = 45 degreesThe 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=3.722Next, we simplify the expression by multiplying the numerator and denominator by 222:v=3.7·222Simplifying further:v=3.7·22Therefore, the puma must leave the ground with a velocity of approximately v=5.24m/s to reach a height of 3.7 m.

Vasquez · Accepted Answer

The vertical motion of the puma can be described by the equation:y=v0sin(θ)·t−12gt2where 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∘)·t−12gt2The horizontal motion of the puma can be described by the equation:x=v0cos(θ)·twhere 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∘)·tSolving the second equation for t, we get:t=0v0cos(45∘)=0Substituting this value of t into the first equation, we can solve for v0:3.7=v0sin(45∘)·0−12g·02=0Therefore, 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 · Accepted Answer

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)2gWhere:- 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)2·9.8Simplifying further, we get:3.7=v2·(22)22·9.8Simplifying and solving for v2, we have:v2=3.7×2·9.8(22)2Solving this equation, we find:v2=72.728Taking the square root of both sides, we get:v=72.728≈8.526m/sTherefore, 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.

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

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