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Math Help - projectile motion

  1. #1
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    projectile motion

    Question:

    An archer hits a target 80m away and the time taken to do this was 1.67 seconds. The target has a radius of 60cm, and its stands at a height of 1.8m, with the 1.8m mark being its centre. What range of angles could have been used for the archer to hit anywhere on the target.

    My attempt:

    v(t) = vcos(theta) + ( vsin(theta) - 10t) [gravity = 10m/s/s]
    s(t) = vcos(theta) *t + ( vsin(theta) *t - 5t^2)


    Distance = 80m, Time = 1.67 seconds

    Upper height of target = 1.8m - 0.6m = 1.2m
    Lower height of target = 1.8m + 0.6m = 2.4m

    I did:

    distance = rate*time
    80m = rate*1.67 seconds
    rate = 48m/s

    then using t = 1.67 seconds, and v = 48m/s

    Lower limit:
    48sin(theta)(1.67) - 5(1.67)^2 = 1.2m

    theta = 10.87 degrees

    Upper limit:
    48sin(theta)(1.67) - 5(1.67)^2 = 2.4m

    theta = 11.79 degrees

    I am not sure if this is the right way to do this question.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Mr.Ree View Post
    Question:

    An archer hits a target 80m away and the time taken to do this was 1.67 seconds. The target has a radius of 60cm, and its stands at a height of 1.8m, with the 1.8m mark being its centre. What range of angles could have been used for the archer to hit anywhere on the target.

    My attempt:

    v(t) = vcos(theta) + ( vsin(theta) - 10t) [gravity = 10m/s/s]
    s(t) = vcos(theta) *t + ( vsin(theta) *t - 5t^2)
    What are these equations supposed to represent, and where did they come from?

    Distance = 80m, Time = 1.67 seconds

    Upper height of target = 1.8m - 0.6m = 1.2m
    Lower height of target = 1.8m + 0.6m = 2.4m

    I did:

    distance = rate*time
    80m = rate*1.67 seconds
    rate = 48m/s

    then using t = 1.67 seconds, and v = 48m/s

    Lower limit:
    48sin(theta)(1.67) - 5(1.67)^2 = 1.2m

    theta = 10.87 degrees

    Upper limit:
    48sin(theta)(1.67) - 5(1.67)^2 = 2.4m

    theta = 11.79 degrees

    I am not sure if this is the right way to do this question.
    This looks like gobbledygook, tell us what you are trying to do and why.

    Also what course is this set as homework for, you have posted it Pre-Algebra and Algebra. If so you must have been given some background information that we don't have.

    CB
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  3. #3
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    We got given it as projectile motion.

    the v(t) equation represents velocity at time 't'
    the s(t) equation represents distance at time 't'

    the cos(theta) components are horizontal, and the sin(theta) components are vertical, and gravity is downwards

    I am trying to find the angles at which the arrow could have been fired in order to hit anywhere on the target. The range of angles.

    since the vertical component of s(t) is the vertical height, i am substituting in time and velocity to find the angles.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Mr.Ree View Post
    We got given it as projectile motion.

    the v(t) equation represents velocity at time 't'
    the s(t) equation represents distance at time 't'

    the cos(theta) components are horizontal, and the sin(theta) components are vertical, and gravity is downwards

    I am trying to find the angles at which the arrow could have been fired in order to hit anywhere on the target. The range of angles.

    since the vertical component of s(t) is the vertical height, i am substituting in time and velocity to find the angles.
    The equations appear to be confounding the vertical and horizontal components of velocity and position and cannot be right.

    Also, it would appear that the archer is standing in a hole.

    CB
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  5. #5
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    I don't understand what I have done incorrectly. How are the equations wrong?

    In regards to the archer, since no information about his position was given, I assumed that he was situated at the origin.

    I am in need of help please.
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