Projectile Motion given 2 position equation

In summary, the conversation discusses modeling the motion of a human body as a particle at the center of mass, and provides equations for the displacement of an athlete's center of mass during a jump. The questions ask for the athlete's position, vector velocity at takeoff, and distance jumped. Solutions are provided for each question, with some confusion over the interpretation of the initial position in question (a).
  • #1
steph35
11
0

Homework Statement


The motion of a human body through space can be precisely modeled as the motion of a particle at the body's center of mass. The components of the displacement of an athlete's center of mass from the beginning to the end of a certain jump are described by the two equations below, where t is the time at which the athlete lands after taking off at time t = 0.

xf =0 + (11.1 m/s)(cos 20.3°)t
0.369 m = 0.812 m + (11.1 m/s)(sin 20.3°)t - 1/2(9.80 m/s2)t2

Homework Equations


(a) Identify his position.
? meters ihat + ? meters jhat
(b) Identify his vector velocity at the takeoff point.
? m/s at ? ° above horizontal
(c) The world long jump record is 8.95 m. How far did the athlete in this problem jump?
? m

The Attempt at a Solution


So, I drew out a diagram and it seemed to make sense to me but the answers were wrong. for a) i got you would just take the component vectors, so Vxi= (11.1 m/s)(cos 20.3) and Vyi= (11.1 m/s) (sin20.3). But that was wrong. I got b). and for c) i thot you could jus pick out the information and it would be .369 but then realized that is only in the y direction, but time isn't give, so how are we supposed to find the x_final?
 
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  • #2
steph35 said:

Homework Statement


The motion of a human body through space can be precisely modeled as the motion of a particle at the body's center of mass. The components of the displacement of an athlete's center of mass from the beginning to the end of a certain jump are described by the two equations below, where t is the time at which the athlete lands after taking off at time t = 0.

xf =0 + (11.1 m/s)(cos 20.3°)t
0.369 m = 0.812 m + (11.1 m/s)(sin 20.3°)t - 1/2(9.80 m/s2)t2

Homework Equations


(a) Identify his position.
? meters ihat + ? meters jhat
(b) Identify his vector velocity at the takeoff point.
? m/s at ? ° above horizontal
(c) The world long jump record is 8.95 m. How far did the athlete in this problem jump?
? m

The Attempt at a Solution


So, I drew out a diagram and it seemed to make sense to me but the answers were wrong. for a) i got you would just take the component vectors, so Vxi= (11.1 m/s)(cos 20.3) and Vyi= (11.1 m/s) (sin20.3). But that was wrong. I got b). and for c) i thot you could jus pick out the information and it would be .369 but then realized that is only in the y direction, but time isn't give, so how are we supposed to find the x_final?

For a) you identified his components of velocity as his position.
But the question asked about the Position P(t) in terms of the unit vectors.

For c) you have two equations in t. Once you solve for t, you can determine x.
 
  • #3
all right, i got the t and solved c.

however, for a) I am still a bit confused. are they asking for his initial position? becuase if they are wouldn't it be 0 for ihat and .443 jhat? i got the 0 ihat correct, but the.443 was wrong?
 
  • #4
steph35 said:
all right, i got the t and solved c.

however, for a) I am still a bit confused. are they asking for his initial position? becuase if they are wouldn't it be 0 for ihat and .443 jhat? i got the 0 ihat correct, but the.443 was wrong?

I'm going to say that I think it is worded poorly. If it is initial position that is asked for, then it will be the x=0 and y=.812 at t=0 The final position is a consequence of his jump and as far as initial position is concerned is not a factor.

My initial reading of the question suggested to me they wanted the unit vectors as a function of the time. But if x=0 is the coefficient of i-hat then I suppose initial condition is what they asked. But I am not entirely happy with that interpretation.
 

What is projectile motion?

Projectile motion is the motion of an object that is thrown or launched into the air and moves along a curved path due to the influence of gravity.

What is the 2-position equation for projectile motion?

The 2-position equation for projectile motion is x = x0 + v0xt + 1/2at2 and y = y0 + v0yt + 1/2at2, where x and y represent the position of the object at time t, x0 and y0 represent the initial position, v0x and v0y represent the initial velocity, and a is the acceleration due to gravity.

How do you solve for the initial velocity in the 2-position equation?

To solve for the initial velocity in the 2-position equation, you can use the equation v0y = gtmax, where g is the acceleration due to gravity and tmax is the time it takes for the object to reach its maximum height.

What is the maximum height reached by a projectile?

The maximum height reached by a projectile can be calculated using the equation hmax = (v0y)2/2g, where v0y is the initial vertical velocity and g is the acceleration due to gravity.

How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object and changing its trajectory. This can make the object travel a shorter distance and reach a lower maximum height compared to a scenario with no air resistance. The effect of air resistance is more significant for objects with larger surface areas and lower velocities.

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