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I\'m working on a problem that gives me the initial velocity andangle at which t

ID: 1764001 • Letter: I

Question

I'm working on a problem that gives me the initial velocity andangle at which the ball is thrown and that it happens on the moon,but that's it. So I only have 2 knowns in the x and y direction.I'm supposed to find how far and how high the ball is thrown beforestriking the ground. Am I able to set the final velocities equal tozero so I can have 3 of 5 variables? The examples I've looked athaven't done that so I'm confused. Is it possible to solvethis?

ok

so

x-component
dx = ?
vx0 = 44.69cos60
vf =
a = 0
t =

y-component
dy = ?
vy0 = 44.69sin60
vf =
a = -g = -1.62m/s2(on the moon)
t =

Explanation / Answer

   MOTION ALONG VERTICAL DIRECTION: (Maximum height )    At max. height, final velocity in verticaldirection, Vyf = 0                           displacement, S = max height (H) = ?           (Vyf ) 2 - ( Vyo ) 2 = 2a S              0 - ( 44.69 sin 60 ) 2 = 2 ( - 1.62 ) H         by solving, H =462.3 m     MOTION ALONG HORIZONTAL DIRECTION: (Horizontal Range )        Vertical displacement during the entiremotion is zero because it comes back to the ground.                   displacement, S = 0                   time of flight, t = T           S = (Vyo) t + ( 1/2 ) at2            0 = ( 44.69 * sin 60 ) T - ( 1/2 ) ( g) T2                T = 2 * 44.69 * sin 60 / 1.62 = 47.78 s          Horizontal Range, R = Horizontal velocity * Time of flight                                          = 44.69 * cos 60 * 47.78 = 1067.64 m                T = 2 * 44.69 * sin 60 / 1.62 = 47.78 s          Horizontal Range, R = Horizontal velocity * Time of flight                                          = 44.69 * cos 60 * 47.78 = 1067.64 m              h