1. Oct 16, 2006

needhelp83

1.) A projectile is shot from the edge of a cliff 125 m above ground level with an initial
velocity of 105 m/s at an angle of 37o with the horizontal.
Determine:
a) the time it takes for the projectile to hit the ground
b) the distance traveled by the projectile as measured from the base of the cliff
c) find the horizontal and vertical components of the projectile's velocity the instant
before hitting the ground
d) the magnitude of the velocity
e) the angle made by the velocity vector with the horizontal.

Horizontal
s = ?
u = 105cos37 = 83.8567 m/s
v = "
a = 0
t = ?

Vertical
s = -125 m
u = 105sin37 = 63.1906 m/s
v = ?
a = -9.8 m/s/s
t = ?

Use:
v2 = u2 + 2as
v2 = (63.1906 m/s)2 + 2(-9.8 m/s/s)(-125 m)
|v| = 80.2686 m/s but v must be going down so
v = -80.2686 m/s

A) v = u + at
-80.2686 m/s = 63.1906 m/s + (-9.8 m/s/s)t
t = 14.6387 s = 14.6 s

B)
vav = s/t
83.8567 m/s = s/14.6387 s
s = 1227.5528 m = 1230 m or 1.23 km

C)
when it hits the ground the velocity components are:
83.8567 m/s x + -80.2686 m/s y or 83.9 m/s x + -80.3 m/s y (c)

D)
This has a magnitude of
(83.85672 + 80.26862)1/2 = 116.08 m/s = 116 m/s (d)

E)
Which forms an angle of
tan-1(80.2686/83.8567) = 43.7o below the horizontal (e)

2. Oct 16, 2006

gunblaze

Yup. All are correct.

Maybe you could perhaps elaborate how u arrive at this answer. The answer is right but the working seems a little wrong.