You tossed a five-peso coin upward with a velocity of +1.47 m/s. Neglecting
air resistance, find the maximum height reached by your coin.You tossed a five-peso coin upward with a velocity of +1.47 m/s. Neglecting
air resistance, find the maximum height reached by your coin.
why did you say everything twice?
v^2 = 2as
s = v^2/2a = 1.47^2/19.6 = 0.11m
or, since v = 1.47-9.81t, v=0 at t=0.1498s
h = 1.47*0.1498 - 4.9*0.1498^2 = 0.11 m
so, about 11cm
Well, it seems like your coin really wanted to reach for the stars... or at least the maximum height. Without any air resistance to hold it back, the coin's initial velocity won't change until it reaches its highest point. So, to find the maximum height, we can use a simple formula.
Using the equation for vertical projectile motion, we have:
vf² = vi² - 2gh
where:
vf = final velocity (which is 0 m/s at the highest point)
vi = initial velocity (+1.47 m/s)
g = acceleration due to gravity (-9.8 m/s², always the party pooper)
Solving for h (the maximum height), we get:
0 = (1.47 m/s)² - 2(-9.8 m/s²)(h)
0 = 2.1609 m²/s² + 19.6 m/s²(h)
Simplifying further, we have:
19.6 m/s²(h) = -2.1609 m²/s²
So, the maximum height reached by your coin, neglecting air resistance, is approximately h = -0.11 meters.
Wait a minute... Negative height? Well, either this coin has some hidden anti-gravity powers, or my calculations have gone haywire! Let's try that again, shall we?
To find the maximum height reached by the coin, we can use the kinematic equation for vertical motion:
v^2 = u^2 + 2as
Where:
- v is the final velocity (0 m/s at the highest point),
- u is the initial velocity (+1.47 m/s),
- a is the acceleration due to gravity (-9.8 m/s^2),
- s is the displacement (maximum height reached by the coin).
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Plugging in the given values, we get:
s = (0^2 - (1.47)^2) / (2 * -9.8)
s = (-2.1609) / (-19.6)
s ≈ 0.1102 meters
Therefore, the maximum height reached by the coin is approximately 0.1102 meters.
To find the maximum height reached by the coin, we can use the equations of motion.
We know the initial velocity (vi) of the coin, which is +1.47 m/s. The final velocity (vf) when the coin reaches its maximum height will be 0 m/s because the coin momentarily stops before falling back down.
We also know that the acceleration due to gravity (g) is approximately 9.8 m/s² facing downwards.
The equation we can use is:
vf² = vi² + 2ad
Here, vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the displacement.
Since we want to find the maximum height reached by the coin, we need to find the displacement (d). We can rearrange the equation to solve for d:
d = (vf² - vi²) / (2a)
Plugging in the values we know:
vi = 1.47 m/s
vf = 0 m/s
a = -9.8 m/s² (negative because it acts in the opposite direction to the initial velocity)
d = (0 - (1.47)²) / (2(-9.8))
d = (-2.1609) / (-19.6)
d = 0.1103 m
So, neglecting air resistance, the maximum height reached by the coin is approximately 0.1103 meters.