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=[tex]v_i=12 m/s[/tex]
The angle at which the dolphin leaves the water is calculated as
[tex]tan \alpha= \frac{h}{d}\\ =\frac{4.10}{5.50} = 0.745\\\alpha=tan^{-1}(0.745) = 36.7\º[/tex]
The x and y components of initial velocity are given as
[tex]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[/tex]
Time to complete the horizontal distance by the dolphin is calculated as
[tex]S=v_i_x \times t\\t=\frac{S}{v_i_x}\\t=\frac{5.5}{9.62}\\t=0.57 s[/tex]
Now the vertical position of ball at this instant is given as
[tex]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[/tex]
Now the location of ball at the instant is given as
[tex]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[/tex]
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.
