Turn on thread page Beta
    • Thread Starter
    Offline

    8
    ReputationRep:
    I won't write out the full question but it basically says a ski jumper of mass 80kg leaves a runway with initial velocity of 20m/s in a horizontal direction and lands after 4.0s.

    1) calculate the horizontal distance travelled in 4.0s

    For my answer I got 80m after multiplying horizontal velocity by time.

    2) calculate the vertical fall of the skier

    This confused me a little but I attempted to do it and my answer came out as 157.0m which seemed slightly off to me but I got this by multiplying the time by g to give me the vertical component then multiplied this by time.

    I'm really lost on how to get the vertical distance. Also can someone please give me tips on how to approach projectile motion questions because I feel like I can do them but I never know which equation to use.

    Posted from TSR Mobile
    Online

    20
    ReputationRep:
    (Original post by TinM)
    I won't write out the full question but it basically says a ski jumper of mass 80kg leaves a runway with initial velocity of 20m/s in a horizontal direction and lands after 4.0s.

    1) calculate the horizontal distance travelled in 4.0s

    For my answer I got 80m after multiplying horizontal velocity by time.

    2) calculate the vertical fall of the skier

    This confused me a little but I attempted to do it and my answer came out as 157.0m which seemed slightly off to me but I got this by multiplying the time by g to give me the vertical component then multiplied this by time.

    I'm really lost on how to get the vertical distance. Also can someone please give me tips on how to approach projectile motion questions because I feel like I can do them but I never know which equation to use.

    Posted from TSR Mobile
    multiplying time by acceleration isn't a valid way of getting a displacement. You need to use one of the suvat equations to get a displacement from a time and an acceleration.

    What you know (vertically)
    u=0
    t=4
    a=g

    What you want to know
    s
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by Joinedup)
    multiplying time by acceleration isn't a valid way of getting a displacement. You need to use one of the suvat equations to get a displacement from a time and an acceleration.

    What you know (vertically)
    u=0
    t=4
    a=g

    What you want to know
    s
    Thank you. I used s=1/2at^2 and came out with 78.5m. I was just wondering do I only need the vertical component of a projectile when I need to find (t)? Because I always get confused about when I should leave the vertical velocity as zero.

    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by TinM)
    Thank you. I used s=1/2at^2 and came out with 78.5m. I was just wondering do I only need the vertical component of a projectile when I need to find (t)? Because I always get confused about when I should leave the vertical velocity as zero.

    Posted from TSR Mobile
    Set vertical to zero!? Nooooo.

    If you have a projectile you separate into the component vertical and horizontal velocities. The horizontal component in normal A level cases has no no force acting on it, so velocity is constant. Therefore... you can't use it to work out time. If it were to never hit the ground it would keep on going at the same horizontal velocity. However, if you consider the vertical component, it does have a force acting on it, therefore an acceleration. Eventually it will hit the ground/the velocity will be come zero at it's peak. So knowing this information you use the vertical component to calculate t.

    You don't set it to zero, it's still there.
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by Oliver.Farren)
    Set vertical to zero!? Nooooo.

    If you have a projectile you separate into the component vertical and horizontal velocities. The horizontal component in normal A level cases has no no force acting on it, so velocity is constant. Therefore... you can't use it to work out time. If it were to never hit the ground it would keep on going at the same horizontal velocity. However, if you consider the vertical component, it does have a force acting on it, therefore an acceleration. Eventually it will hit the ground/the velocity will be come zero at it's peak. So knowing this information you use the vertical component to calculate t.

    You don't set it to zero, it's still there.
    Okay, thank that's cleared up a lot of confusion

    Posted from TSR Mobile
 
 
 
Poll
Have you ever experienced bullying?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.