A ball is thrown upwards . Its height varies with time as shown in figure. If the acceleration due to gravity is `7.5 m//s^(2)`, then the height h is

Options

A. 10m
B. 15m
C. 20m
D. 25m

Answer

B

Explanation

The height h is convred in time interval t=1s to t =2s or t=5s to t=6s.
Let u be the initial velocity given to the ball when projected vertically upwards. Then
`h=s_(2)-s_(1)=s_(5)-s_(6)` .....(i)
`s_(2)-s_(1)=(uxx2-(1)/(2)xx7.5xx2^(2))-(uxx1-(1)/(2)xx7.5xx1^(2))`
`=u-(3)/(2)xx7.5` .....(ii)
`s_(5)-s_(6)=(uxx5-(1)/(2)xx7.5xx5^(2))-(uxx6-(1)/(2)xx7.5xx6^(2))`
`=-u+(11)/(2)xx7.5`
`therefore u-(3)/(2)xx7.5=-u+(11)/(2)xx7.5`
or `2u=(14)/(2)xx7.57xx7.5`
or `2u=(14)/(2)xx7.5=7xx7.5`
or `u=7xx(7.5)/(2)=26.25 m//s`
`therefore h=26.25-(3)/(2)xx7.5=26.25-11.25=15.0m`