Establish the eqaution for position time relation (i.e., `s = ut + (1)/(2)at^(2)`) using velocity time graph. How is the equation modified when
(i) a body is just dropped from some height, (ii) a body is thrown vertically upwards with some velocity ?

Calculate the maximum height attained by the body thrown vertically upward with a velocity of 9.8m//s (g=9.8m//s^2)

The distance (s) travelled by a body in time (t) is given by s = ut + (1)/(2)at^(2) . How is the formula modified when (i) body starts from rest , (ii) motion of body is uniform.

A body is thrown vertically upwards which graph represents variation of velocity w.r.t time

Using velocity time graph, establish the relation s = ut + (1)/(2)at^(2) , where the symbols have their usual meanings.

a. A ball is thrown vertically upwards. Aftre some time it trturns to the throuer. Draw the velocity-time graph and speed-time graph. b. A ball is dropped from some height. After rebounding from the floor, it ascends to the same height. Draw the velocity-time graph and speed-time graph.

From the velocity-time graph of uniformaccelerated motion deduce the equations of motion in (i) velocity and time (ii) distance and time (iii) disstance and velocity.

A body is projected vertically upward from the surface of the earth, then the velocity-time graph is:-

Indicate the velocity (v) - time (t) graphs for a body: (i) falling under gravity. (ii) thrown vertically upwards till it falls back to the ground after reaching the highest point.

Derive the following equations of motion for uniformly accelerated motion from velocity-time graph : s = ut + 1/2 at^(2)