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Determining the height of an angled projectile after 10m

  1. Aug 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Jason is practicing his tennis stroke by hitting balls against a wall. The ball leaves his racquet at a height of 43 cm above the ground at an angle of 80° with respect to the vertical. The speed of the ball as it leaves the racquet is 25 m/s and it must travel a distance of 10 m before it reaches the wall. How far above the ground does the ball strike the wall?

    2. Relevant equations
    I'm not really sure? Vertical displacement probably.
    Vertical displacement = voy (time) + (1/2)(acceleration)(time)2

    3. The attempt at a solution
    I (think I) found the initial velocities in the vertical and horizontal directions:

    vinitialx = vinitial cos angle = 25cos(80) = 4 m/s
    vinitialy = vinitial sin angle = 25sin(80) = 25 m/s

    With that, I attempted to find how long it takes the ball to reach the wall

    10m = 4 m/s(t)
    2.5 m/s = t

    That's where I get lost. I try taking the displacement formula and plugging in the numbers and then adding 0.43 meters to the answer, but none of my answers are correct. I've tried both adding and subtracting (1/2)(9.81)^2 from 25(2.5), because I'm not really sure which it would be here, but both are wrong. Where am I messing up?
     
  2. jcsd
  3. Aug 27, 2015 #2

    billy_joule

    User Avatar
    Science Advisor

    You've taken the angle as 80deg with respect to the horizontal. The question statement says 80 deg wrt the vertical.

    Also, you shouldn't round here prematurely:

    Best to work symbolically then do a single calculation at the end rather than accumulate rounding errors and increase the chance of calculation errors.
     
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