A rock is thrown straight up with an initial speed of 8.00 m/s from the roof of a building 12.0m above the ground. For the motion from the roof to the ground, what are the magnitude and the direction of:
(a) the average velocity of the rock?
(b) the average acceleration of the rock?
2007-07-23 16:09:11 · 4 answers · asked by Imran 1 in Science & Mathematics ➔ Physics
a)
The average velocity only depends on the initial and final positions of the rock (that is, the roof and the ground).
Find the total time of the flight:
You know: v0 (= 8.0 m/s), d (= -12.0 m), a (= -9.8 m/s²)
You want: t
So use this equation: d = v0·t + (1/2)·a·t²
(1/2)a·t² + v0·t - d = 0
t = [-v0 ± √(v0² + 2a·d)] / a = [-8 ± √(64 + 235.2)] / -9.8
= 2.58 s
v(avg) = d / t = -8 m / 2.58 s = -3.1 m/s
b)
a(avg) = -g = -9.8 m/s²
2007-07-23 16:23:31 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The average accleration of the rock is -9.8 m/sec^2.
The average velocity is defined as (1/t) INT( v dt) where t goes from 0 to the time at the end of the trip.
Given a = -9.8, v =vo+at. or v = 8-9.8*t
INT v dt = INT [8 - 9.8 t ] = 8*t - 4.9 t^2 + k.
When t=0, k=12
Now we have to find the time at which the stone hits the ground. This is when 4.9t^2 - 8*t - 12 =0
(I multiplied everything by -1). From quad formula,
t = [8+/- sqrt(64 - 4*4.9*(-12) ] / 9.8 =
[ 8 +/- sqrt(300) ] /9.8
only the positive root is realistic and is 10sqrt(3)= 17.3 appx. So the final t = 25.3/9.8 = 2.6 sec appx.
The average velocity is -12 m /2.6 sec = - 4.7 m/sec appx.
2007-07-23 16:43:16 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
a) ave speed = initial speed(u) + final speed(v)/ 2, but v = u +at where a = acceleration of free fall and t = time
in this case time can be found using this equation
g = 2h/t^2
where g = acceleraion of free fall (9.8ms^-2)
h = height 12 m
t = time
so 9.8 = 24/t^2
t = sqrt(24/9.8)
t = 1.57s
final speed (v) is found by this equation v = u + at
where v = fnal speed
u = initial speed (8m/s)
a = acceleration of free fall (9.8m/s^2)
t = time (1.57s)
you get v = 8 + (9.8 x 1.57)
v = 8 + 15,4
v = 23.4m/s
now applying the formula v(ave) = v + u/2
where v = final speed u = initial speed
you get v = 23.4 + 8/2
v = 15.7m/s veritically downwards
b) The acceleration of any object free falling near the erth's surface (neglecting friction) is contant at 9.81m/s^2 acting vertically downwards toward the earth's center
2007-07-23 16:37:00 · answer #3 · answered by Mandél M 3 · 0⤊ 0⤋
This is the only thing I remember from physics:
Acceleration = Accel. due to gravity = -9.8 m/s^2
2007-07-23 16:35:37 · answer #4 · answered by Ipsulis 3 · 0⤊ 0⤋