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In this problem, you will apply kinematic equations to a jumping flea.

problem 1) 

In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/s2 . Ignore air resistance.

1. A flea jumps straight up to a maximum height of 0.510m. What is its initial velocity as it leaves the ground?
3.16m/s

2. How long is the flea in the air from the time it jumps to the time it hits the ground?
time in air=____s


Problem 2) 

A rock is thrown vertically upward with a speed of 19.0 {rm m/s} from the roof of a building that is 50.0 m above the ground. Assume free fall.1. In how many seconds after being thrown does the rock strike the ground?t=___s2. 

What is the speed of the rock just before it strikes the ground?


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Answer #1
Concept and reason

Use the concept of equations of motion, which includes the initial speed, final speed, acceleration due to gravity, time, and displacement covered to solve this problem.

Use the formula for maximum height reached to calculate the initial velocity. Then, use the formula for time of flight to calculate the time of flight. Use the equation of motion that relates displacement, acceleration, and time to calculate the time. Then use the equation of motion that relates initial and final velocities and time to calculate the velocity of the rock on reaching the ground.

Fundamentals

Displacement is defined as the smallest distance between two points.

Velocity is defined as the rate of change of displacement.

Acceleration is defined as the rate of change of velocity.

The relation between the initial velocity, final velocity, acceleration, and the displacement covered is,

v2u2=2as{v^2} - {u^2} = 2as

Here, u is the initial velocity, v is the final velocity, a is the acceleration, and s is the displacement.

The displacement is given by,

s=ut+12at2s = ut + \frac{1}{2}a{t^2}

Here, u is the initial velocity, a is the acceleration, and t is the time.

The velocity of the rock on reaching the ground is,

v=ugtv = u - gt

Here, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

The maximum height reached by an object projected vertically upwards is,

h=u22gh = \frac{{{u^2}}}{{2g}}

Here, u is the initial velocity and g is the acceleration due to gravity.

The total time for which the object remains in air is called time of flight.

The total time of flight is given by,

T=2ugT = \frac{{2u}}{g}

Here, u is the initial velocity and g is the acceleration due to gravity.

(1-1)

The maximum height reached by the flea is,

h=u22gh = \frac{{{u^2}}}{{2g}}

Rearrange the equation for the initial velocity.

u=2ghu = \sqrt {2gh}

Substitute 9.80m/s29.80{\rm{ m/}}{{\rm{s}}^2} for g and 0.510m0.510{\rm{ m}} for h.

u=2(9.80m/s2)(0.510m)=3.16m/s\begin{array}{c}\\u = \sqrt {2\left( {9.80{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {0.510{\rm{ m}}} \right)} \\\\ = 3.16{\rm{ m/s}}\\\end{array}

(1-2)

The total time of flight of the flea is,

T=2ugT = \frac{{2u}}{g}

Here, u is the initial velocity and g is the acceleration due to gravity.

Substitute 3.16m/s3.16{\rm{ m/s}} for u and 9.80m/s29.80{\rm{ m/}}{{\rm{s}}^2} for g.

T=2(3.16m/s)9.80m/s2=0.645s\begin{array}{c}\\T = \frac{{2\left( {3.16{\rm{ m/s}}} \right)}}{{9.80{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 0.645{\rm{ s}}\\\end{array}

(2-1)

The displacement is given by,

s=ut+12at2s = ut + \frac{1}{2}a{t^2}

Here, u is the initial velocity, a is the acceleration, and t is the time.

The rock falls down so the displacement is negative. The acceleration of the rock is due to gravity.

Substitute h - h for s and g - g for a.

h=ut+12(g)t2gt22ut2h=0\begin{array}{c}\\ - h = ut + \frac{1}{2}\left( { - g} \right){t^2}\\\\g{t^2} - 2ut - 2h = 0\\\end{array}

Substitute 19.0m/s19.0{\rm{ m/s}} for u, 9.80m/s29.80{\rm{ m/}}{{\rm{s}}^2} for g, and 50.0m50.0{\rm{ m}} for h.

(9.80m/s2)t22(19.0m/s)t2(50.0m)=0t=5.675s\begin{array}{c}\\\left( {9.80{\rm{ m/}}{{\rm{s}}^2}} \right){t^2} - 2\left( {19.0{\rm{ m/s}}} \right)t - 2\left( {50.0{\rm{ m}}} \right) = 0\\\\t = 5.675{\rm{ s}}\\\end{array}

Round off to three significant figures. So, the time taken by the rock to reach the ground is5.68s5.68{\rm{ s}}.

(2-2)

The velocity of the rock on reaching the ground is,

v=ugtv = u - gt

Substitute 19.0m/s19.0{\rm{ m/s}} for u, 9.80m/s29.80{\rm{ m/}}{{\rm{s}}^2} for g, and 5.675s5.675{\rm{ s}} for t.

v=(19.0m/s)(9.80m/s2)(5.675s)=36.6m/s\begin{array}{c}\\v = \left( {19.0{\rm{ m/s}}} \right) - \left( {9.80{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {5.675{\rm{ s}}} \right)\\\\ = - 36.6{\rm{ m/s}}\\\end{array}

The negative sign indicates that the rock is falling down. The magnitude of velocity of the rock on reaching the ground is36.6m/s36.6{\rm{ m/s}}.

Ans: Part 1-1

The velocity with which the flea leaves the ground is3.16m/s3.16{\rm{ m/s}}.

Part 1-2

The flea is in the air for0.645s0.645{\rm{ s}}.

Part 2-1

The time after which the rock reaches the ground is5.68s5.68{\rm{ s}}.

Part 2-2

The magnitude of velocity of the rock on reaching the ground is36.6m/s36.6{\rm{ m/s}}.

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