Question
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.
Answers
Elena
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.
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.