A coin is tossed straight up. Find:
a.) The initial velocity when it is released and reaches the height of 0.75m above where it is thrown.
b.) The final velocity as you catch it from where it is released.
c.) Time it stays in the air.

Question

anonymous · Answer

I wanna answer a. Just a review. :)

Given: dy= 0.75 m          g= 9.8 m/s^2
          Vf = 0

Formula: 2gdy= Vf^2- Vi^2

2(9.8)(0.75)= 0^2- Vi^2
14.5- 0= Vi^2
$$- \sqrt{14.5}= Vi^2$$ 
3.83 m/s= Vi^2

anonymous · Answer

Lol fail. ** 3.83 m/s = Vi

anonymous · Answer

answer also b and c, celina tahahahaha :))

anonymous · Answer

Blaaah D: 
Given: dy= 0.75 
m g= 9.8 m/s^2
 Vi = 0 

Formula: 
2gdy= Vi^2- Vf^2 
2(9.8)(0.75)= 0^2- Vf^2 
14.5- 0= Vf^2 
√14.5=Vf2 
3.83 m/s= Vf

anonymous · Answer

c. Formula:

t=vf-vi/ g

Substitute:

t= (0-3.83)/-9.8
t= 0.39 s

anonymous · Answer

the answer for final velocity is incorrect :P