Position as a function of time

  • Thread starter jti3066
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  • #1
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Homework Statement



Need someone to check my work please (this is a review for my final)

The position vector r(t) ( in m) of a football as a function of time t (in s) is given by:

r(t) = 6t + (8t-4.9(t^2))[j]


A) Calculate the intial velocity of the football (at t=0) in the x and y coordinates (ie..Cartesian coordinates)

B)How long does it take for the football to reach max height?

C) What is the velocity ( give magnitude and direction) at the max height?

D) What is the horizontal distance traveled by the fooball before it hits the ground?


Homework Equations



v = 6 + (8 - 9.8t)[j]......(firt derivative of position vector)

v_f = v_o + at

x_f = x_i + v_o(t) + .5(a)(t^2)

The Attempt at a Solution



A) r(t)' = 6 + (8 - 9.8t)[j]


when t = 0

r(0) = 6 + 8[j]

so

v = sqrt( ^2 + [j]^2) = sqrt( 6^2 + 8^2) = 10 m/s

B) At max height the velocity will be zero so:

v_f = v_o + at

0 = (10) + (-9.81)t

t = -10/-9.81 = 1.01 s

C) At max height the velocity will be zero

D) The horizontal distance traveled by the ball will be:

x_f = x_o + v_ot + .5(a)(t^2)

x_f = 0 + (10)(2.02) + .5(9.81)(2.02^2) = 40.03 m
 
Last edited:

Answers and Replies

  • #2
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For part c) I think I found a mistake in my work...The velocity will not be zero in the x direction but will be in the y direction...so my aswer should be:

V_f = 10 + (-9.91)(1.01) = 0.092 m/s (at max height)

Is this correct?
 
  • #3
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Part B is incorrect.

The overall velocity is 10, but to reach max height, you're only concern is with v_y - like you mentioned in your last post.

Thus, your numbers for C and D are off, too.

For D, what is the acceleration in the x direction?
 
  • #4
46
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Ok so here are the new answers:

B) V_fy = V_oy + at

0 = 8 + (-9.81)(t)

t = 0.815 s

C) V_f = 10 + (-9.81)(0.815) = 2.00 m/s

D) x_f = 0 + 10(1.63) + .5(0)(1.63^2)

x_f = 16.3 m
 
  • #5
136
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B is correct, now - but you didn't break down the velocity into x- and y- components for parts C and D.

C should be very easy, as you've already defined that the velocity in the y- direction goes to zero at max height. So, what does that leave you with?

And for D, you basically did it right the first time, with the exceptions that you used overall velocity, and you used the incorrect acceleration.
 
  • #6
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ok i'm confused now...

C) 0 m/s

D) would I use: a = 0 and v_0= 6
 
  • #7
136
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You're on target for D, now.

But the ball doesn't stop when it reaches max height. What is it doing?
 
  • #8
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Is O m/s correct for part C....

Part D: I know how the ball travels through the air but I am having trouble relating the info to the equation.........

x_f = 0 + (6)(1.63) + .5(-9.81)(1.63^2)

what about know......if not please give me the correct input please
 
  • #9
46
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thanks for your help p21bass........

Would someone mind chech the answers in post 8 please...TIA
 

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