Answer :
a) The reaction force of the ground is 2.5mg
b) The velocity upon takeoff is 3.7 m/s
c) The initial angle of takeoff is [tex]20.3^{\circ}[/tex]
d) The distance travelled is 7.6 m
Explanation:
a)
We can solve this part by applying Newton's third law of motion, which states that:
"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"
In this situation, we can identify the person as object A and the ground as force B. Therefore:
- The person exerts a force of 2.5mg on the ground (action)
- As a result, the ground also exerts a force of 2.5mg (in the opposite direction) on the person (reaction)
So, the force exerted by the ground is
[tex]F=2.5 mg[/tex]
b)
For this part, we just need to analyze the situation along the vertical direction, since the force is applied along the vertical direction.
The impulse exerted on the person is equal to the change in momentum of the person along this direction, so we can write:
[tex]F\Delta t = m(v-u)[/tex]
where the term on the left is the impulse while the term on the right is the change in momentum, and where we have
F is the net force applied on the person
t = 0.25 s is the time during which the force is applied
m is the mass of the person
u = 0 is the initial velocity in the y-direction
v is the final velocity in the y-direction after the 0.25 s
The net force applied on the person is equal to the force applied on the ground minus the weight of the person (mg), therefore:
F = 2.5mg - mg = 1.5mg
Solving for v, we find the vertical component of the velocity upon takeoff:
[tex]v=\frac{F\Delta t}{m}=\frac{(1.5mg)(0.25)}{m}=3.7 m/s[/tex]
c)
At take-off, we know that the components of the velocity are:
[tex]v_x = 10 m/s[/tex] is the horizontal component
[tex]v_y = 3.7 m/s[/tex] is the vertical component
The initial angle of takeoff is given by
[tex]\theta = tan^{-1}(\frac{v_y}{v_x})[/tex]
And substituting, we find
[tex]\theta = tan^{-1}(\frac{3.7}{10})=20.3^{\circ}[/tex]
above the horizontal.
d)
The motion of the person is the same as the motion of a projectile, so the horizontal range will be given by the equation
[tex]d=\frac{u^2 sin(2\theta)}{g}[/tex]
where
u is the initial speed
[tex]\theta=20.3^{\circ}[/tex] is the angle of projection
g = 9.8 m/s^2 is the acceleration of gravity
Here we need to find the initial speed first, which is the magnitude of the resultant of horizontal and vertical component of the initial velocity:
[tex]u=\sqrt{v_x^2+v_y^2}=\sqrt{10^2+3.7^2}=10.7 m/s[/tex]
And now, by substituting, we find:
[tex]d=\frac{(10.7)^2(sin (2\cdot 20.3))}{9.8}=7.6 m[/tex]
Learn more about Newton's third law of motion and projectile motion:
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