Question

A dolphin jumps with an initial velocity of 15 m/s at an angle of 45 degrees above the horizontal. The dolphin passes through the center of a hoop before returning to the water. The dolphin is moving horizontally when it goes through the hoop.

a) What are the x and y components of the initial velocity?
b) What are the acceleration in horizontal and vertical direction?
c) How high above the water is the center of the hoop?

Answer 1

distance=5.74 m

Step-by-step explanation:

velocity components:

`v_x=15* cos45=(15\sqrt2)/(2)\\ v_y=15* sin45=(15\sqrt2)/(2)`

acceleration components

`x-axis=0 m/s^2\\y-axis=g=-9.8m/s^2`

since the dolphin is moving horizontally going through the center of the hoop, we can assume that the vertical velocity=0

We can thus find the distance required to reach the hoop by kinematics:

`v_y^2=u_y^2+2ad`

`0=((15\sqrt2)/(2))^2+2* (-9.80)* d\\ 19.6d=112.5\\d=5.74m`(Roughly)

Or you can use the formula for maximum height of a body (in this case the dolphin) undergoing projectile motion:

`h_m_a_x=(u^2sin^2(\theta))/(2g)=(15^2* sin^2(45))/(2* 9.8)\\=5.74 (roughly)`