The angle that dolphin leaves the water:
tanα = h/d =4.10/5.50 = 0.745,
α =arctan0.745 = 36.7º.
The x- and y- components of the initial velocity:
v(ox) = v(o) •cosα = 12•cos 36.7º = 9.62 m/s,
v(oy) = v(o) •sin α = 12• sin 36.7º = 7.17 m/s,
v(x)=v(ox)
x = v(x) •t,
t = x/ v(x) = 5.5/9.62 = 0.572 s.
y = v(oy) •t - g•t²/2 = 7.17•0.572 – 9.8•(0.572)²/2 = 2.5 m.
The ball’s location at this instant of time:
y = y(o) +v(o)t - g•t²/2 =
= 4.1 + 0•0.572 - 9.8•(0.572)²/2 = 2.5 m/s.
A trained dolphin leaps from the water with an initial speed of 12 m/s. It jumps directly towards a ball held by the trainer a horizontal distance of 5.50 m away and a vertical distance of 4.10 m above the water. In the absence of gravity the dolphin would move in a straight line to the ball and catch it, but because of gravity the dolphin follows a parabolic path well below the ball’s initial position. If the trainer releases the ball the instant the dolphin leaves the water, show that the dolphin and the falling ball meet.
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