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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?

v = 6

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)

A) r(t)' = 6v_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) = 6when t = 0

r(0) = 6

*+ 8[j]*

so

v = sqrt(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 mB) 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

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