Question

Example: Leap of FaithA dolphin leaps from the water at an initial speed of 12.0 m/s toward a ball held by a trainer a horizontal distance of 5.50 m away. If the trainer drops the ball the instant the dolphin leaves the water, show that the dolphin and the ball meet

Answer 1

Answer:

The location of both ball and the dolphin is same i.e 2.5 m above at the time 0.57 seconds this indicates that the ball and dolphin will meet.

Explanation:

As the complete question is not visible, therefore, the question is searched online and following reference question is obtained and is attached as the attachment.

From the reference question data is as follows

initial speed of the dolphin=$v_i=12 m/s$

The angle at which the  dolphin leaves the water is calculated as

$tan \alpha= \frac{h}{d}\\ =\frac{4.10}{5.50} = 0.745\\\alpha=tan^{-1}(0.745) = 36.7\º$

The x and y components of initial velocity are given as

$v_i_x=v_i cos \alpha=12*cos 36.7\º = 9.62 m/s\\v_i_y=v_i sin \alpha=12*sin 36.7\º = 7.17m/s$

Time to complete the horizontal distance by the dolphin is calculated as

$S=v_i_x \times t\\t=\frac{S}{v_i_x}\\t=\frac{5.5}{9.62}\\t=0.57 s$

Now the vertical position of ball at this instant is given as

$y_{dolphin} = v_i_y \times t-g\frac{t^2}{2}\\y_{dolphin}= 7.7 \times 0.57-10\frac{0.57^2}{2}\\y_{dolphin}=2.5 m$

Now the location of ball at the instant is given as

$y_{ball}=y_{ball-o}+v_i_y_{ball} \times t-g\frac{t^2}{2}\\y_{ball}=4.1+0 \times 0.57-10\frac{0.57^2}{2}\\y_{ball}=2.5 m$

As the location of both ball and the dolphin is same i.e 2.5 m above at the time 0.57 seconds this indicates that the ball and dolphin will meet.