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Homework Help: SUVAT projectile motion

  1. Dec 29, 2015 #1
    1. The problem statement, all variables and given/known data
    A golfer is about to hit a golf ball from a tee to a hole on the green. The tee and the green are at the same level. The golfer hits the ball, projecting it at a speed of 35 m/s 28° to the horizontal. Air resistance is negligible. Calculate:
    a) the horizontal distance travelled by the ball before first landing on the green (2 marks)
    b) the maximum height reached by the ball above the level of the tee (2 marks)

    2. Relevant equations

    3. The attempt at a solution
    a) On the horizontal plane:
    s = ?
    u = 35cos28 = 30.9 m/s
    v = ?
    a = 0
    t = ?

    Okay, let's see if we can find v. With the horizontal component of the velocity remaining constant, we'd need to work out the vertical component and resolve. But I'd need the time to work out the final vertical component with v = u + at, which is useless, because I'd be able to use d = st if I had the time. I'm not seeing an obvious way to find it.

    EDIT: Hold up - I'm just looking at the horizontal component here, and that's going to remain constant, so v = u = 30.9 m/s. But where would I go from here? Could I just double the value for time I found in part b), since the maximum height would be at the half point of the trajectory? (Right?)
    1.67 x 2 = 3.34s
    3.34 x 30.9 = 103.2m. Anyone?

    b) On the vertical plane:
    s = ?
    u = 35sin28 = 16.4 m/s
    v = 0
    a = -9.8
    t = (0 - 16.4) / -9.8 = 1.67s

    s = (16.4 x 1.67) + (0.5 x -9.8 x 1.672) = 13.7m. Can someone confirm?
    Last edited: Dec 29, 2015
  2. jcsd
  3. Dec 29, 2015 #2
    What does SUVAT mean?
  4. Dec 29, 2015 #3
    This going somewhere? It's a set of projectile motion equations.
  5. Dec 29, 2015 #4
    Yes, right, you could also use:

    t= u/g

    h= u^2 /g - 1/2 *g* (u^2/g^2)= u^2/(2g)

    h= u^2 /(2g)

    (It's the same thing but I express the maximum h using u and g)
  6. Dec 29, 2015 #5
    What does it stand for?

    Saying the relevant equations for a projectile motion problem is "a set of projectile motion equations" does not really answer the question.
  7. Dec 29, 2015 #6
    This seems tedious, but I'll play along.
    s = displacement
    u = initial velocity
    v = final velocity
    a = acceleration
    t = time

    Are you making a point that s = displacement, not distance?

    EDIT: I've edited in a proposed solution to a). Could you take a look?
  8. Dec 29, 2015 #7


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    You appear to have calculated the time for the vertical velocity to reach zero. That's not the time for the whole flight.

    Once you have the total flight time that will be the same for the horizontal motion which is constant velocity.
  9. Dec 29, 2015 #8


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    Oops I miss read your post. You did remember to double the time.
  10. Dec 30, 2015 #9


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    looks good.
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