A projectile is shot from the edge of a cliff h = 265 m above ground level with an initial speed of v0 = 105 m/s at an angle of 37.0° with the horizontal(adsbygoogle = window.adsbygoogle || []).push({});

Xo=0m Yo=0m

X= 1361m YTop of Cliff=265m

Ax= 0m/s/s Ay=-9.8m/s/s

Vo= 105 m/s Vx= 83.86 m/s

Theta= 37.0

T= 16.23 s

I need to determine the vertical component of the velocity at the point directly above the ground before impact. I calculated the vertical velocity from launch to the maximum Y to be 63.2 m/s.

I tried calculating this several ways but I know all of the following are wrong:

v=105+(-9.8)(16.23)= -54

V^2=105^2+2(-9.8)(0-265)= 127.35

V^2=0^2+2(-9.8)(0-265)= 72.07

I know I'm missing something here but I cant quite put my finger on it. Could someone direct me to what ever this is?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Tricky Projectile Motion

**Physics Forums | Science Articles, Homework Help, Discussion**