Question

1. A trained dolphin leaps from the water with an initial speed of 12.0 m/s. It jumps directly toward 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 at the instant the dolphin leaves the water, show that the dolphin and the falling ball meet. h= 4.10 m d 5.50 m

Answer 1

Given,

vi = 12 m/s ; D = 5.5 m ; y = 4.1 m ;

the angle with which dolphin projects herself

theta = tan^-1 (4.1/5.5) = 36.7 deg

vx = 12 cos36.7 = 9.62 m/s

vy = 12 sin36.7 = 7.17 m/s

we know that, D = vx t => t = D/vx

t = 5.5/9.62 = 0.572 s

We know that,

Y = Y0 + vy t + 1/2 at^2

Y = 0 + 7.17 x .572 - 0.5 x 9.8 x .572^2 = 2.5 m

At this time, the ball is at (using the same eqn)

Y(ball) = 4.1 + 0 - 0.5 x 9.8 x 0.572 = 2.5 m

Hence, Y(dolphin) = Y(ball) = 2.5; Hence they meet.