i do not understand how to solve this equation. imstuck. a 2.05m tall b ball pla

Question

i do not understand how to solve this equation. imstuck. a 2.05m tall b ball player takes a shot when he is 6.02m fromthe basket. if the launch andgle is 25.0 degree's and the ball waslaunched at the level at the players head, what must be the releasespeed of the ball for the playyer to make the shot? the basket is35.0m above the floor. i do not understand how to solve this equation. imstuck. a 2.05m tall b ball player takes a shot when he is 6.02m fromthe basket. if the launch andgle is 25.0 degree's and the ball waslaunched at the level at the players head, what must be the releasespeed of the ball for the playyer to make the shot? the basket is35.0m above the floor.

Explanation / Answer

   the angle of projection is 25.0o
   the components of the initial velocity are givenby
   vix = vi cos25o
        = vi(0.906)
   viy = vi sin40o
        = vi(0.422)
   now the time taken by the ball to move 10.0 mhorizontally is
   t = (x / vix)
   = (10.0 m) / (vix)
   so at this time the vertical displacement of the ballmust be
   y = (35.0 m - 2.00 m)
   = 33.0 m
   now using the equation of motion we get the initialspeed as
   y = viy t + (1 / 2) ayt2
   33.0 m = (vi (0.422)) ((10.0 m) /vi (0.906)) + (1 / 2) (-9.80 m / s2) ((10.0m) / vi (0.906))2
   vi = ....... m / s

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