<p>A projectile is shot from the edge of a cliff h = 125 m above ground level wi

Question

<p>A projectile is shot from the edge of a cliff h = 125 m above ground level with an initial speed of v0 = 145 m/s at an angle of 37.0&#176; with the horizontal. <br /><br />At the instant just before the projectile hits point the ground, find the horizontal and the vertical components of its velocity.<br />What is the vertical velocity in m/s?<br />What is the horizontal velocity in m/s?<br />What is the magnitude of the velocity? <br />What is the angle made by the velocity vector with the horizontal?</p>

Explanation / Answer

so if the angle with the ground is

then:

vix=vocos

viy=vosin

to find final velocity in the y:

first find the maximum height by using:

v2-vi2=2ad where a=-g, v=0 b/c projectile is at peak, and vi is initial v in y direction:

-(vosin)2/-2g=d   note that this d is height traveled above INITIAL height, which is 125 m

d=388.51 m

peak=d+hcliff=514 m

now use the same equation for the falling portion of it's motion, except vi is zero

vf=(2ahpeak)

vf=-100 m/s note this is the y component and it is in the negative direction.

there is no acceleration in the x direction, so

vfx=vix=vocos=116 m/s

V=(vfx2 +vfy2)

V=153 m/s

Direction can be found using

tan=vfy/vfx

=-40.8 degrees   as in the velocity is directed at and angle of 40.8 degrees below the horizontal

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