Question

A baseball pitcher throws a pitch horizontally from a height of 6 feet with an initial speed of 130 feet per second. a) find a vector valued function describing the position of the ball t seconds after release. b) if home plate is 60 feet away, how high is the ball when it crosses home plate? c)if a person drops a ball from height of 6 feet at the same time the pitcher releases the ball, how high will the dropped ball be when the pitch crosses home plate?

Answer 1

Let's take horizontal direction, of projection of ball as +ve x direction and vertical downward as +ve y direction and point of projection as origin, Initial velocity of ball
v_i = 130 i + 0 j
Acceleration of ball is g down ward , that is 0 i +g j
Position vector of ball after time t,
r(t) = v_i t + a t² /2
      = 130t i + 32.17 t² /2 j

b) As ball crosses home plate 60 ft away, x coordinate of ball = 60
     130 t = 60
t = 6/13
y coordinate = 32.17*(6/13)² /2 = 3.43 ft
Hence ball is 3.43 ft below the point of projection.
Height of ball = 6-3.43 = 2.57 ft

c) Ball dropped and ball projected have same initial velocity in vertical direction, that is zero, projected from same height and both move with acceleration , equal to g in vertical direction. Hence both have same vertcal position at any instant of time.
Hence ball dropped will be at height of 2.57 ft when pitch crosses home plate.