A student throws a coin vertically downward from the top of a building. The coin leaves the thrower’s hand with a speed of 15.0 m/s. How far does it fall in 2.00 s?
Δh = [1/2 * (-9.81 m/s^2) * (2.00 s)^2] - (15.0 m/s * 2.00 m)
height = -4.905t^2 - 15t + k
at t = 0, height = k
at t = 2.00 , height = -4.905(4) - 15(2) + k = -49.62 + k
change in height in those 2 seconds = k - (-49.62 + k) m
= 49.62 m
V = Vo+g*T = 15+9.8*2 = 34.6 m/s.
V^2 = Vo^2+2g*d = (34.6)^2
15^2+19.6d = 1197.2
d = 49.6 m.
To find the distance the coin falls in 2.00 seconds, we can use the equation for the distance traveled by an object in free fall:
d = v₀t + (1/2)at²
Where:
d = distance
v₀ = initial velocity (speed)
t = time
a = acceleration due to gravity
In this case, we are given that the initial velocity (v₀) is 15.0 m/s and the time (t) is 2.00 seconds.
The acceleration due to gravity (a) is a constant value of approximately 9.8 m/s² when the object is near the surface of the Earth.
Plugging in the values into the equation:
d = (15.0 m/s)(2.00 s) + (1/2)(9.8 m/s²)(2.00 s)²
Simplifying the equation:
d = 30 m + (1/2)(9.8 m/s²)(4.00 s²)
d = 30 m + (1/2)(9.8 m/s²)(16.00 s²)
d = 30 m + 78.4 m
d = 108.4 m
Therefore, the coin falls a distance of 108.4 meters in 2.00 seconds.