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?
3 answers
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
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.
V^2 = Vo^2+2g*d = (34.6)^2
15^2+19.6d = 1197.2
d = 49.6 m.
5 answers