A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is `10 ms^-1` , then the maximum height attained by the stone is (`g= 10 ms^(-2)`)

Answer

(A)-r, (B)-p, (C)-r (D)-s

Explanation

At `t=0, v(0)=10 m//s, t=0, v(6)=0`
Change `v(6)-v(0), Deltav=0-10=-10 m//s`

Average acceleration
`=("charge in velocity")/("time")=(-10)/6=(-5)/3 m//s^(2)`
Average velocity `=("Displacement")/("time interval")`
Total displacement = Area of `Delta's` (with +ve or -ve)
`=1/2xx2xx10-1/2xx4xx10=-10 m` (units)
`:.` Average velocity `=(-10)/6=-(-5)/3 m//s`
`a(3)=` slope of line which exist at `t=0 I_(0) t=4`
`a= tan theta=(-10)/2 =-5`