a boy tosses a coin upward with a velocity of 14.7 m/s.

Question

a. find the maximum height reached by the coin.
b. time of flight
c. velocity when the coin returns to the hand
d. suppose the boy failed to catch the coin and the coin goes to the ground, with what velocity will it strike the ground?the boy's hand is 0.49 m above the ground.

Answer 1

v=u-gt or 0=14.7-9.81t
t=14.7/9.81=1.498 say1.5 sec
v^2=u^2-2gh or 0=14.7^2-2x9.81xh
h=14.7^2/2x9.81=11m.
Add 0.49m Total ht=11.49m
On falling down, u=0
v^2=0+2x9.81x11 or v=14.69 upto hand.
v^2=0+2x9.81x11.49 or v=15.01m/s up to ground.

Answer 2

a) Use the formula; V^2=Vo^2=2g(y-yo)
cancel V^2 and yo (because it is zero)
-V0^2/2g=2gy/2g <-- Cancel 2g
y=-Vo/2g=-(14.7 m/s2)/2(-9.8m/s^2)
y=11.04m

b)t=2y/Vo=2(11.04m)/14.7 m/s
t=1.5s
t^2(1.5)(2)
=3s

c)V^2=(2g)(yo)
=2(-9.8 m/s^2)(-11.04)
V^2= <SQUARE-ROOT>216.38
=-14.71 m/s

Answer 3

g = -9.8 m/s^2
Vo = 14.7 m/s V = 0 m/s
Yo = 0 m Y = 11.04 m
To = 1.5 s T = 1.5 s

a.) (V-Vo)/2g = (Y-Yo)
y= (-216.09)/(-19.16)
y= 11.04 m
b.) V= Vo+gt
0= 14.7 + (-9.8)(t)
-14.7 = -9.8t
t= 1.5 s
total time = (1.5)(2)
total time = 3 s
c.) V= Vo+gt
V= 0+(-9.8)(1.5)
V= 14.7 m/s
~~~~~~~~~~~~~~~~~~~~~~~~~
Yo = 11.04 + 0.49
Yo = 11.53 m
Vo = 0 m/s
V = ???
t = ???
g = -9.8

d.) V^2 = Vo^2 + 2g(Y-Yo)
V^2 = 0 + 2(-9.8)(0-11.53)
V^2 = (-19.6)(-11.53)
V^2 = 225.988
V = 15.03 m/s^2
~~~~~~~~~~~~~~~~~~~~~~~~~
I'm not really sure about this one. The answer should be negative because of the downward velocity. But if the value was negative it would be imaginary because of the radical sign.

Yo = 0 ; Y = 11.53

V^2 = 0 + 2(-9.8)(11.53-0)
V^2 = (-19.6)(11.53)
(sqrt)V^2 = (sqrt)(-225.988)
V = 15.03i m/s
~~~~~~~~~~~~~~~~~~~~~~~~~

Answer 4

Correction

d.) V^2 = Vo^2 + 2g(Y-Yo)
V^2 = 0 + 2(-9.8)(0-11.53)
V^2 = (-19.6)(-11.53)
V^2 = 225.988
V = 15.03 m/s <---

Answer 5

What does the Yo represents?