Answer:
A. [tex]v_{x}=21.4m/s[/tex]
[tex]v_{y}=26.42m/s[/tex]
B. [tex]t=2.7s[/tex]
C. [tex]y_{max} =35.61m[/tex]
D. [tex]x=115.34m[/tex]
E. [tex]a_{x}=0[/tex]
[tex]a_{y}=-9.8m/s^2=g[/tex]
F. [tex]v_{y}=0[/tex]
[tex]v_{x}=21.4m/s[/tex]
Explanation:
From the exercise our initial values are:
[tex]v_{o}=34m/s[/tex]
[tex]\alpha=51º[/tex]
A. The horizontal and vertical components are:
[tex]v_{x}=34cos(51)=21.4m/s[/tex]
[tex]v_{y}=34sin(51)=26.42m/s[/tex]
B. At maximum height the y-component of velocity becomes 0
[tex]v_{y}=v_{o}+a_{y}t[/tex]
[tex]0=26.42-9.8m/s^2*t[/tex]
[tex]t=2.7s[/tex]
C. The maximum height above the ground is:
[tex]v_{y} ^{2}=v_{o}^2+2a_{y}(y-y_{o})[/tex]
At maximum height the y-component of velocity becomes 0
[tex]0=(26.42)^2-2(9.8)y[/tex]
[tex]y=\frac{-(26.42)^2}{-2(9.8)}=35.61m[/tex]
D. To find how dar from its firing point does the sell land we need to calculate how much time does it take to do it first
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^2[/tex]
When the shell land y=0
[tex]0=0+(26.42)t-\frac{1}{2}(9.8)t^2[/tex]
Solving the quadratic equation for t
[tex]t=0[/tex] or [tex]t=5.39s[/tex]
Since time can not be 0 t=5.39s
[tex]x=v_{ox}t=(21.4m/s)(5.39s)=115.34m[/tex]
E. Since the velocity at the horizontal component is constant
[tex]a_{x}=0[/tex]
The vertical acceleration of the shell is gravity
[tex]g=-9.8m/s^2[/tex]
F. At highest point the vertical component is 0. The shell stops going up ans start to go down
[tex]v_{y}=0[/tex]
[tex]v_{x}=v_{oy}+a_{x}t[/tex]
Since [tex]a_{x}=0[/tex]
[tex]v_{x}=21.4m/s[/tex]
