Respuesta :
The magnitude of the vertical velocity under gravity changes by approximately 9.81 m/s each second, The option that correctly relates the magnitude, , of the vertical velocity is the option: (vy1=vy1)≥vy3.
The speed of rocks = u
The height the rocks are thrown = h₀
Rock 1; Direction = horizontal above the 45°
Rock 2; Direction =horizontal below45°
Rock 3; Direction = 0° or in the horizontal direction
The vertical component of Rock 1= -u·sin(45°)
Vertical component of Rock 2 = u·sin(45°)
Vertical component of Rock 3 = u·sin(0°) = 0
The velocity of the rock
v² = u² - 2·g·h₀
v= ±√u2-2.g.h0
v= ±√u2-2.g.h0
Therefore, the vertical velocity the rocks, Rock 1 and Rock 2, thrown with magnitude of velocity,|vy|=|u.sin45°| is the same during the vertical motion of the ball at the same height, but change only in sign
at given height h°, vy1=vy2
therefore , before the rocks hit the ground , vy1=vy2
the initial vertical velocity of rocks3, vy3=0,therefore:
∴ vy3=±√v+2.g.h°≤√u²+2.g.h°= (vy1= vy1
there fore , the correct option , we have:
vy1=vy1≥vy3
three identical rocks are launched with identical speeds from the top of a platform of height 0 h . the rocks are launched in the directions indicated above. question which of the following correctly relates the magnitude vy of the vertical component of the velocity of each rock immediately before it hits the ground?
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