Question

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.

Answer 1

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

Answer 2

Lol fail. ** 3.83 m/s = Vi

Answer 3

answer also b and c, celina tahahahaha :))

Answer 4

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

Answer 5

c. Formula:

t=vf-vi/ g

Substitute:

t= (0-3.83)/-9.8
t= 0.39 s

Answer 6

the answer for final velocity is incorrect :P