A rock is dropped from a rooftop. It takes 10 seconds to hit the ground. what was the height the rock was dropped from in meters(ignore air resistance).
2007-06-16 06:48:05 · 4 answers · asked by Danny 4 in Science & Mathematics ➔ Physics
Xf = .5at^2 + Vi*t + Xi
Xf = final position
a = acceleration (-9.8m/s^2)
t = time
Vi = initial velocity
Xi = initial position
We are looking for the initial postion of the rock. When the rock hits the ground, is final position is 0m. Because the rock is dropped from rest, is initial velocity is 0m/s
0 = .5(-9.8)(10)^2 + 0(10) + Xi
0 = -490 + Xi
Xi = 490m
2007-06-16 06:56:16 · answer #1 · answered by 7 · 1⤊ 0⤋
S = ut + 1/2 at^2; where u = 0 the initial velocity at t = 0 and a = g = 9.81 m/sec^2 the acceleration due to gravity on Earth's surface. S is the distance traveled in t = 10 seconds. You now have the equation and numbers to solve for S, the height of the rock when it was dropped.
2007-06-16 15:26:16 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
S = ut +1/2 a t^2.
u = 0
a = g = 9/8 m/s^2
t = 10 .
S = 1/2 x 9.8 x 100 = 490 m
2007-06-16 15:17:23 · answer #3 · answered by Pearlsawme 7 · 0⤊ 0⤋
gravity is 32 feet per second per second
2007-06-16 14:00:56 · answer #4 · answered by Manz 5 · 0⤊ 0⤋