Answer:
Rachel(2.5,0)
ball(6.5,4.7)
b.R=10.15m/s, 27.57deg
Step-by-step explanation:
The reference angle of Rachel is 0
![0^(0) while the ball is at 36^(0)](https://img.qammunity.org/2020/formulas/physics/high-school/4p731y6fpol73xmc5mi76fofesagsdj6j6.png)
resolving rachel's speed to the horizontal, we have
Ux=2.5cos0
Ux=2.5m/s
resolving rachel's speed to the vertical we have,
Uy=2.5sin0
Uy=0
for the ball
resolving the speed to its horizontal component
Ux=8cos36
Ux=6.5m/s
Uy=8sin36
Uy=4.7m/s
Rachel(2.5,0)
ball(6.5,4.7)
To get the resultant of their speed
Add the horizontal speed of rachel to that of the ball to get the total horizontal speed
Add the vertical speed of rachel and the ball to get the total vertical speed component
TUx=2.5+6.5=9M/S
TUy=0+4.7=4.7m/s
R=
![√((TUx^2+TUy^2)](https://img.qammunity.org/2020/formulas/physics/high-school/csedj0dq82toqyen8w9n8q0j1thshn7q4t.png)
R=
![√((9^2+4.7^2)](https://img.qammunity.org/2020/formulas/physics/high-school/fegh2sufuo8wirmwmipxoyl47pb6v9ar7x.png)
R=
![√((103))](https://img.qammunity.org/2020/formulas/physics/high-school/1auy3pq4nk0bh2pakejh8mgekxv5ycu7fq.png)
R=10.15m/s
the direction
tan
=TUy/TUx
tan
=4.7/9
=tan^-1(0.522)
=27.57deg