Howard J. answered • 11/04/19
Principal Mech Engr with 35+ years' on-the-job physics experience
a projectile is launched from ground level to the top of a cliff which is 135m tall. The projectile is shot with an initial velocity of 75m/s at an angle of 60 degrees with the horizontal.
a) draw a diagram of the projectiles motion. include the initial velocity vector(Vo) the initial velocity vectors in the x and y directions (Vox and Voy), the initial and final heights (Yo and Yf), the horizontal distance (X1) and the projectile's path
I'll let you do this.
b) Calculate the x and y components of the projectiles initial velocity. Add these values to your diagram.
Vox=75cos60=38 m/sec
Voy=75sin60=65 m/sec
c) Determine the time for the projectile to reach its landing.
H=Voyt-0.5gt2=135 m
65t-4.9t2=135
4.9t2-65t+135=0
t2-13.26t+27.55=0
t={13.26±√[(13.26)2-4*27.55]}/2={13.26±√65.62)}/2={13.26±8.1}/2= 6.6±4.0
t=10.6 sec or t=2.6 sec
The apex of the trajectory is when Vy=0
Vy=Voy-gt=0
t=Voy/g=65/9.81=6.6 sec (no surprise there)
The projectile reaches a 135 m height twice: once on the way up then once on landing.
t=10.6 sec
d) determine the horizontal distance the projectile from the base of the cliff
X1=(Vox)t=38(10.6)=403 m
e) at the instant just before the projectile hits its landing zone at the top of the cliff, determine the horizontal and vertical components of its velocity.
V=(Vox)i + (Voy-gt)j=38i + (65-9.91t)j
when it lands, t=10.6 sec
V=38i-39j m/sec
f) at the instant just before it strikes the ground, what is the angle below the horizontal made by the velocity vector.
tanB=|-39/38|
B=46º
g) determine the maximum height above the launch zone reached by the projectile.
we found t=6.6 sec at the apex
H=Voyt-4.9t2=65(6.6)-4.9(6.6)2=215 m
