Posted by **Anonymous** on Wednesday, September 26, 2007 at 8:55pm.

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 -
**bobpursley**, Wednesday, September 26, 2007 at 9:09pm
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) -
**Anonymous**, Wednesday, September 26, 2007 at 10:25pm
would the angle of impact be -60 degrees?

vx=vx0=18m/s

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

Theta=-60

- physics -
**bobpursley**, Thursday, September 27, 2007 at 7:05am
Yes, theta is -60 deg if your 31m/s is correct (I didn't check that).

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