An object is projected vertically upward from the ground. Show that it takes the object the same amount of time to reach its maximum height as it...
takes for it to drop from that height back to the ground. Measure height in meters.
There are no measurements or numbers or units of any kind to work with.
Let u be the initial velocity of the object thrown up verically.
So the ovject goes on loosing its velocity till the final velocity of the object becomes zero.
Therefore v = u-gt = 0, where v is the final velocity and g is the acceleration due to the gravity. So u = gt or time when the velocity becomes zero is u/g.
Therefore the height reached by the object h = ut -(1/2)gt^2. Substituting t = u/g in this equation h = u(u/g)-(1/2)g(u/g)^2 = u^2/g-u^2/2g = u^2/2g. Therefore the maximum height reached by the object is u^2/2g in time t = u/g.
Now we calculate the time required for the object to fall from the height u^2/2g to the ground . Let t be the time to reach the ground.
Then the equation of motion is h = u^2/g = u't+(1/2)gt^2, where u' = is the initial velocity = 0 when the object has reached the highest point and lost all its velocity and begins fall.
u^2/2g = 0*t +(1/2)gt^2 . We solve for t. We multiply both sides by g/2:
(2/g) *u^2/2g = t^2.
u^2/g^2 = t^2.
Therefore u/g = t. Or the time for the object to reach the ground from the highest point is u/g which is the same as the time to reach the highest point from the ground.