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-0.5gt^{2}=135 m

65t-4.9t^{2}=135

4.9t^{2}-65t+135=0

t^{2}-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=**38**i **+ (65-9.91t)**j**

when it lands, t=10.6 sec

**V**=38**i**-39**j **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-4.9t^{2}=65(6.6)-4.9(6.6)^{2}=**215 m**