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

Answer

B

Explanation

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