Respuesta :
Answer:193.90 m/s
Explanation:
Given
launch angle [tex]\theta =31.7^{\circ}[/tex]
launch velocity [tex]v_0=202 m/s[/tex]
Horizontal velocity of the shell after [tex]t=18.96 s[/tex]
time of flight of Projectile [tex]T=\frac{2u\sin \theta }{g}[/tex]
[tex]T={\frac2\times 202\sin 31.7}{9.8}[/tex]
[tex]T=21.66 s[/tex]
i.e. projectile is declining as [tex]t>\frac{T}{2}[/tex]
but horizontal component of velocity will remain same as there is no opposing force in horizontal direction
Horizontal component of velocity is
[tex]u_x=v_0\cos \theta =202\cdot \cos 31.7=193.90 m/s[/tex]
Answer:
171.86 m/s
Explanation:
vo = 202 m/s
α = 31.7°
t = 18.96 s
In case of projectile motion, the acceleration along the horizontal direction is zero. It means the magnitude of velocity along horizontal direction is always constant. The horizontal component of velocity is given by
vx = vo Cosα
vx = 202 Cos 31.7
vx = 171.86 m/s
Thus, the horizontal component of velocity is 171.86 m/s.
