Answer:
a)[tex]S=2.57[/tex]
b)[tex]H=203.2yard[/tex]
c)[tex]X=121.79yards[/tex]
Step-by-step explanation:
From the question we are told that:
Angle [tex]\theta=1.2=68.755^o[/tex]
a)
Generally the equation for Slope is mathematically given by
[tex]S=tan \theta[/tex]
[tex]S=tan 68.755[/tex]
[tex]S=2.57[/tex]
b)
Given the right Angle triangle with horizontal distance [tex]x=85yard[/tex]
Generally the equation for Height traveled is mathematically given by
[tex]H=tan\theta*x[/tex]
[tex]H=2.57*85[/tex]
[tex]H=218.45[/tex]
[tex]H=203.2yard[/tex]
c)
Generally the equation for Horizontal Distance traveled at 313 height traveled is mathematically given by
[tex]X=\frac{313}{tan65.75}[/tex]
[tex]X=\frac{313}{2.57}[/tex]
[tex]X=121.79yards[/tex]
Using the projectile principle, the slope of the rockets path, height above the ground and the distance traveled at a height of 313 yards would be 2.57, 218.63, 121.69 respectively.
Given the Parameters :
Converting to degree :
A.)
The slope of the rocket's path :
Slope = tan(68.755) = 2.57
B.)
Horizontal distance, = distance along the x-axis = 85 yards
Vertical distance = height = distance along y-axis, y
y = slope × x
y = tan(68.755) × 85
y = 218.63 yards
C.)
Vertical distance, y = 313 yards
From :
[tex] y = x tan\theta [/tex]
[tex] x = \frac{y}{tan\theta} [/tex]
[tex] x = \frac{313}{tan(68.755°)} [/tex]
x = 121.69 yards
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