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physics

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A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18m/s. The cliff is 50m above a flat horizontal beach, as shown in Figure 3.20. How long after being released does the stone strike the beach below the cliff? With what speed and angle of impact does it land?

Work:

-50m=-1/2(9.8m/s^2)t^2
t=3.19s

x=(18m/s)(3.19s)=57.4m
vy=0-(9.8m/s^2)(3.19s)=-31.3m/s

Am I doing everything correctly? How do I find the angle of impact?

  • physics -

    correct. Angle of impact comes from the velocity vectors at impact.

    Measuring angle from the beach to the incoming projectile,

    TanTheta=verticalvelocty/horizontalvelocty

  • physics ( for bobpursley) -

    would the angle of impact be -60 degrees?

    vx=vx0=18m/s

    tan Theta=(-31m/s)/(18m/s)
    Theta=-60

  • physics -

    Yes, theta is -60 deg if your 31m/s is correct (I didn't check that).

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