An object is thrown vertically upwards. It rises to a height of 4 m determine
(a) the time to reach the maximum height,
(b) the initial velocity
(c) the velocity 1.0 s after leaving the hand,
(d) the distance traveled vertically during the 1.0 seconds
PLEASE HELP ME, I've got a physics test tomorrow and am in desperate need of help.
(My teacher has a disability I have a feeling were all goig to fail...)
(a) the time to reach the maximum height,
(b) the initial velocity
(c) the velocity 1.0 s after leaving the hand,
(d) the distance traveled vertically during the 1.0 seconds
PLEASE HELP ME, I've got a physics test tomorrow and am in desperate need of help.
(My teacher has a disability I have a feeling were all goig to fail...)
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The key to kinematics problems like this is understanding what information you have. There are 5 quantities in play:
1) initial velocity, or v sub i
2) final velocity, or v sub f
3) acceleration, or a (always constant in these problems)
4) time, or t
5) distance traveled, or d
Each of the "big four" kinematic equations has 4 of these quantities but is missing one. So you generally need three of them to figure out a 4th.
In this case you know: the distance (4m), the acceleration (g or 9.8 meters per seconds squared, the acceleration of earth's gravity), and the final velocity (0, since any object thrown upward has zero velocity for the instant at the top of its "jump").
So, just start solving equations. It's easiest to start with part b).
vf^2 = vi^2 + 2ad
0 = vi^2 + 2(g*4m)
vi^2 = -8m * (-9.8m/s^2) [g = negative 9.8m/s^2 in this case because down is negative]
vi^2 = 784m^2/s^2
vi = 28 m/s
so b) is 28m/s
a) use this equation:
vf = vi + at
0 = 28m/s - 9.8m/s^2(t)
9.8m/s^2 * t = 28m/s
t = 2.86s
so a) is 2.86 s
c) vf = vi + at
vf = 28m/s - 9.8m/s^2 * 1s
vf = 28m/s - 9.8m/s
vf = 18.2m/s
d) (just to throw one more equation at you)
d = (vi + vf)/2 * t
d = (28m/s + 18.2m/s)/2 * 1s
d = 46.2m/s * 1s
d = 46.2m
Hope this helps.
EDIT: I'd just like to point out that defining "down" as negative is arbitrary; it could just as easily be positive. But then throwing an object "up" would be negative so other quantities would have to be reversed: for instance the initial velocity would be -28m/s instead of +28m/s
1) initial velocity, or v sub i
2) final velocity, or v sub f
3) acceleration, or a (always constant in these problems)
4) time, or t
5) distance traveled, or d
Each of the "big four" kinematic equations has 4 of these quantities but is missing one. So you generally need three of them to figure out a 4th.
In this case you know: the distance (4m), the acceleration (g or 9.8 meters per seconds squared, the acceleration of earth's gravity), and the final velocity (0, since any object thrown upward has zero velocity for the instant at the top of its "jump").
So, just start solving equations. It's easiest to start with part b).
vf^2 = vi^2 + 2ad
0 = vi^2 + 2(g*4m)
vi^2 = -8m * (-9.8m/s^2) [g = negative 9.8m/s^2 in this case because down is negative]
vi^2 = 784m^2/s^2
vi = 28 m/s
so b) is 28m/s
a) use this equation:
vf = vi + at
0 = 28m/s - 9.8m/s^2(t)
9.8m/s^2 * t = 28m/s
t = 2.86s
so a) is 2.86 s
c) vf = vi + at
vf = 28m/s - 9.8m/s^2 * 1s
vf = 28m/s - 9.8m/s
vf = 18.2m/s
d) (just to throw one more equation at you)
d = (vi + vf)/2 * t
d = (28m/s + 18.2m/s)/2 * 1s
d = 46.2m/s * 1s
d = 46.2m
Hope this helps.
EDIT: I'd just like to point out that defining "down" as negative is arbitrary; it could just as easily be positive. But then throwing an object "up" would be negative so other quantities would have to be reversed: for instance the initial velocity would be -28m/s instead of +28m/s