A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is observed to be `v = 7.6 hati + 6.1 hatj` in metre per second (`hati` is horizontal, `hatj` is upward). Give the approximate answers.
(a) To what maximum height does the ball rise?
(b) What total horizontal distance does the ball travel?
(c) What are the magnitude and
(d) What are the direction of the ball's velocity just before it hits the ground?

Answer

A, B, C, D

Explanation

`v_x = u_x = 7.6m//s`
`v_y = (sqrt (u_(y)^2 - 2gh))`
`:. u_y = (sqrt(v_(y)^2 + 2gh))`
`= (sqrt((6.1)^2 + 2 xx 10 xx 9.1))`
`= 14.8 m//s`
(a) `H = (u_(y)^2)/(2g) = (14.8)^2/(2xx10) ~~ 11m`
(b) `R = (2u_x u_y)/g = (2xx 7.6 xx 14.8)/10 ~~ 23 m`
(c ) `u= (sqrt(u_(x)^2 + u_(y)^2))`
` = (sqrt (7.6)^2 + (14.8)^2)`
`=16.6 m//s`
(d) `theta = tan^-1 (u_y/u_x)`
`=(tan^-1)(14.8)/(7.6) ~~ tan^-1(2),`
below horizontal.