The height of a building is 122.5 cm. but this time, a stone is thrown upward with a velocity of 10m/s. What is the velocity just before it hits the ground? How long does it take the stone to reach the ground?
Please include the solutions so that I will compute it if it's correct
2006-09-11 04:07:22 · 1 answers · asked by Anonymous in Education & Reference ➔ Homework Help
Given:
d = 122.5 cm.
Vi = 10m/s
2006-09-11 04:13:20 · update #1
Torricelli's equation to find final velocity, given initial velocity, acceleration, and distance:
v1 is positive (up)
a is negative (gravity goes down)
d is negative (endpoint is below the beginning point)
v2 = v1 + 2ad
v2 = 10 m/s + 2 * -9.8 m/s^2 * -1.225 m
v2 = 10 + -24.01
v2 = -14.01 m/s
To find t:
d = v1t + 1/2 at^2
-1.225 = 10t + -4.9t^2
0 = -4.9t^2 + 10t + 1.225
Quadratic formula: (-b +/- (b^2 - 4ac)^1/2)/2a
t = (-10 +/- (100 + 24.01)^1/2)/-9.8
t = 1.02 +/- (-1.136)
t = -0.116, 2.156 s
Ignore the negative solution: t = 2.156s.
2006-09-11 04:59:33 · answer #1 · answered by ³√carthagebrujah 6 · 0⤊ 0⤋