From the top of a toward of height `60 m`, a ball is thrown vertically upward with speed `20 m//s`. After some time, the ball strikes the ground. Sketch the v-t and s-t graph.

Explanation

Assume `O` as the origin and upward direction +ve.
`O` to B throught `A`:
Displacement `s=-60 m`
`s=ut-1/2 g t^(2)`
`-60=20 t-5 t^(2)`
`t^(2)-4t-12=0`
`(t-6)(t+2)=0`
`t=6 s`
The ball will strike the ground after time `t =6 s`.
v-t graph: `v=u-g t`
`v=20-10 t`
`t=0, v=20 m//s`
`t=6s, v=-40 m//s`
`v=0implies20-10timplies t=2 s`
Shape: `y=20-10x` (straight line)
`y=c+mx`
`c=20`,+ve
`m=-10`, -ve

s-t graph: `s=ut-1/2 g t^(2)=20t-5 t^(2)`
`t=0, s=0`
`t=6 s, s=-60 m`
`s=0implies20t-5t^(2)=0impliest(20-5t)impliest=0, t=4 s`
s will be maximum, when `v(=ds//dt)=0implies t=2 s`
At `t=2 s, s=20 m`
A part from the initial and final values, we required values at critical points, where change takes place.
Shape: `y=20x-5x^(2)`(parabola)
The coefficient of `x^(2)` is -ve, i.e. the parabola will be open downward.