Posted by **abebe** on Friday, June 15, 2012 at 11:45am.

a base ball player hits a home run and the ball lands in the left field seats 10m above the point at which the ball was hits. the ball lands with a velocity of 20m/s at angle of 30 degere below the horizontal.ignoring air reference find the initial velocity with with the ball leave the bat.

- physics -
**Elena**, Friday, June 15, 2012 at 5:02pm
Without air RESISTANCE(!), the horizontal component of the velocity is constant

v(x) =v(ox) = v•cos30 = 20•cos30 = 17.32 m/s.

The path of the projectile is symmetrical respectively the max height, therefore , there is the point("left" point) on the ascending branch of parabolic trajectory at the height h= 10 m.

The magnitude of the vertical component of velocity v(y) at the “left” point is equal to the magnitude v(y) at the “right” point (their directions are opposite). Vertical component of the initial velocity and vertical component of velocity at height 10 m are related as

v(oy)² - v(y)² = 2•g•h, =>

v(oy) =sqrt {2•g•h+v(y)²} =

= sqrt {2•g•h+(v•sin30)²} =

= sqrt {2•9.8•10+(20•sin30)²} =17.2 m/s.

v(o) =sqrt{v(ox)²+v(oy)²} =

= sqrt{17.32²+ 17.2²} = 24.4 m/s

tan α =v(oy)/v(ox) = 17.2/17.32=0.99

α =44.8º

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