Vertical Motion
Vertical motion:
1. When a body is dropped from a height under gravity and neglecting the resistive forces, then, initial velocity of the body (u) is zero. The acceleration of the body is a = g.
The velocity of the body at time t is v = gt [using v = u + at].
The distance travelled by the body for time t is h = 
[using 
].
The velocity of the body when h height is travelled v = 
[using v2 = u2 + 2as].
2. When a body is thrown vertical upward with initial velocity u under gravitational acceleration g, then its final velocity is zero and it travels a maximum height H.
The velocity of the body at time t is v = u – gt [using v = u + at and a = – g].
The distance travelled by the body for time t is h = ut – [using and a = – g].
The velocity of the body when h height is travelled 
[using v2 = u2 + 2as and a = – g].
The maximum height travelled by the body is H = 
[using v2 = u2 + 2as where v = 0 and a = – g].
Time of ascent and time of descent are same: A body is projected vertically upwards with velocity u and after time t it reaches the maximum height H before stop. Then using the equation v = u – gt we get, 0 = u – gt or, t = 
.
When the body is at height H then using the equation v2 = u2 – 2gH
We get, 0 = u2 – 2gH or, H = .
If the body takes time T to reach the ground from maximum height, then using the equation
Or, H = 0.T + 
Or, =
Or, T = .
So, Time of ascent = time of descent.
Example: A juggler juggles balls throwing upwards. He throws one ball when the previous one is at a height halfway of its journey. How high do the balls rise if he throws n balls per second?
Let us consider juggler juggles balls throwing upwards with speed u and maximum height travelled by the ball is h.
Time interval between two balls is 
s. so, 
= u() – 
g()2 or, = 
– 
or, u = 
+ 
At maximum height v = 0 using relation v2 = u2 – 2as we get, h = 
= 
.
Or, 2gh = ( + )2
Or, 8gh = h2n2 + 2gh + 
Or, 6gh = h2n2 +
Or, h2n4 – 6ghn2 + g2 = 0
h = 
= 
= 
.
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