a boy tosses a coin upward with a velocity of 14.7 m/s.
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.
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
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.
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
~~~~~~~~~~~~~~~~~~~~~~~~~
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 <---
What does the Yo represents?
a. Why did the coin have to go up in the first place? Was it trying to reach for the stars? Anyway, to find the maximum height reached by the coin, we need to use some physics magic. The formula you can use is H = (v^2) / (2g), where H is the maximum height, v is the initial velocity of the coin, and g is the acceleration due to gravity. So, plug and chug: H = (14.7^2) / (2 * 9.8). Don't worry, the answer won't make your head spin!
b. Time of flight? Well, that's just a fancy way of asking how long the coin will stay up in the air. We can use the formula t = 2v / g, where t is the time of flight, v is the initial velocity of the coin, and g is the acceleration due to gravity. So, plug and juggle: t = 2 * 14.7 / 9.8. Keep your eyes on the clock!
c. The velocity when the coin returns to the hand? Clearly, the coin wants a high-five from the boy's hand after its travel adventure. Well, to find this velocity, we can just use the initial velocity, which is 14.7 m/s. Talk about a quick reunion!
d. Uh-oh, the hand fumbled the catch? No worries, gravity will ensure the coin's safe landing (or maybe not so safe). To find the velocity with which the coin strikes the ground, we first need to find the time it takes to reach the ground. We can use the formula t = sqrt(2h / g), where h is the height above the ground and g is the acceleration due to gravity. So, t = sqrt(2 * 0.49 / 9.8). And once we know the time, we can multiply it by the acceleration due to gravity to find the velocity. Just keep your fingers crossed for a soft landing!
To solve this problem, we can use the equations of motion. Let's break down each part of the question:
a. To find the maximum height reached by the coin, we need to use the equation for vertical displacement:
Δy = V₀yt - 0.5gt²
Where:
Δy = vertical displacement or maximum height
V₀y = initial vertical velocity (14.7 m/s)
g = acceleration due to gravity (-9.8 m/s²)
t = time (unknown)
Rearranging the equation:
0 = (14.7 m/s)t - 0.5(9.8 m/s²)t²
This is a quadratic equation. We need to solve for t.
0.5(9.8 m/s²)t² - (14.7 m/s)t = 0
Factoring out t:
t(0.5(9.8 m/s²)t - 14.7 m/s) = 0
This equation has two solutions: t = 0 (initial time) and t = (14.7 m/s) / (0.5(9.8 m/s²)).
The maximum height is reached when the coin is at its peak, so we can ignore the t = 0 solution. Therefore, the maximum height can be found using:
Δy = (14.7 m/s) * t - 0.5 * (9.8 m/s²) * t²
b. To find the time of flight, we can use the equation:
Δy = V₀yt - 0.5gt²
We already have the value for the maximum height (Δy) and initial vertical velocity (V₀y), which is 14.7 m/s. The unknown is still t. We can plug in the values and solve for t.
Δy = (14.7 m/s) * t - 0.5 * (9.8 m/s²) * t²
c. To find the velocity when the coin returns to the hand, we can use the equation for vertical velocity:
Vy = V₀y - gt
Plug in the value for V₀y (14.7 m/s) and solve for t.
Vy = 14.7 m/s - 9.8 m/s² * t
d. To find the velocity at which the coin strikes the ground, we need to consider two parts: the upward motion and the downward motion. The time it takes for the coin to reach the ground can be determined by using the equation for time of flight (part b).
Once we have the time, we can find the velocity at which the coin strikes the ground using the equation for vertical velocity:
Vfinal = V₀y - gt
Plug in the time and values for V₀y (14.7 m/s) and g (-9.8 m/s²), then solve for Vfinal.