A major-league pitcher can throw a baseball in excess of 42.7m/s. If a ball is t

Question

A major-league pitcher can throw a baseball in excess of 42.7m/s. If a ball is thrown horizontally at this speed, how much willit drop by the time it reaches a catcher who is 17.3 m away fromthe point of release?
I solved for the time using the information for the xcomponents. When the question states that the ball is thrownhorizontally at a speed of 42.7 m/s, does that mean that that isthe value for the x component of the velocity?I'm at a loss as towhere to go from there.I assume that you have to find the ycomponent of velocity, and eventually use that information to finddisplacement in the y direction. However, I cant seem to compose anequation to solve for the y component of velocity. Please explainthoroughly, thanks!
A major-league pitcher can throw a baseball in excess of 42.7m/s. If a ball is thrown horizontally at this speed, how much willit drop by the time it reaches a catcher who is 17.3 m away fromthe point of release?
I solved for the time using the information for the xcomponents. When the question states that the ball is thrownhorizontally at a speed of 42.7 m/s, does that mean that that isthe value for the x component of the velocity?I'm at a loss as towhere to go from there.I assume that you have to find the ycomponent of velocity, and eventually use that information to finddisplacement in the y direction. However, I cant seem to compose anequation to solve for the y component of velocity. Please explainthoroughly, thanks!

Explanation / Answer

      THIS IS THE CASE OFHORIZONTAL PROJECTION OF A PROJECTILE       Initial Horizontalvelocity, U = 42.7 m/s       Horizontal displacement, X= 17.3 m       Displacement in horizontaldirection, X = U t                                                                   t= X / U = 17.3 / 42.7 = 0.405 s       Displacement in verticaldirection, Y = - (1/2) gt2= - (1/2) (9.8) (0.405)2 = - 0.804 m ( downward)

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