Asked by Anonymous
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?
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?
Answers
Answered by
bobpursley
correct. Angle of impact comes from the velocity vectors at impact.
Measuring angle from the beach to the incoming projectile,
TanTheta=verticalvelocty/horizontalvelocty
Measuring angle from the beach to the incoming projectile,
TanTheta=verticalvelocty/horizontalvelocty
Answered by
Anonymous
would the angle of impact be -60 degrees?
vx=vx0=18m/s
tan Theta=(-31m/s)/(18m/s)
Theta=-60
vx=vx0=18m/s
tan Theta=(-31m/s)/(18m/s)
Theta=-60
Answered by
bobpursley
Yes, theta is -60 deg if your 31m/s is correct (I didn't check that).
Answered by
MARK BEN
I would say to him to fall down a cliff
Answered by
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