Answer:
[tex]v_{o}=141.51m/s[/tex]
Explanation:
From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity
[tex]x=1420m\\g=-9.8m/s^{2}[/tex]
From the equation of x-position we know that
[tex]x=v_{ox}t=v_{o}cos(35)t[/tex]
Solving for [tex]v_{o}[/tex]
[tex]v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)}[/tex] (1)
Now, if we analyze the equation of y-position we got
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
At the end of the motion y=0
[tex]0=v_{o}sin(35)t+\frac{1}{2}gt^{2}[/tex]
Knowing the equation for [tex]v_{o}[/tex] in (1)
[tex]0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}[/tex]
[tex]\frac{1}{2}(9.8)t^{2}=1420tan(35)[/tex]
Solving for t
[tex]t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s[/tex]
Now, we can solve (1)
[tex]v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s[/tex]
Answer:
121.7 m/s
Explanation:
Horizontal range, R = 1420 m
Angle of projection, θ = 35°
acceleration due to gravity, g = 9.8 m/s^2
let u be the velocity of projection.
Use the formula for horizontal range
[tex]R=\frac{u^{2}Sin2\theta }{g}[/tex]
[tex]1420=\frac{u^{2}Sin70 }{9.8}[/tex]
u = 121.7 m/s
Thus, the velocity of projection is 121.7 m/s.
