A ball is tossed from the window of a building with an initial velocity of `8 ms^-1` at an angle of `20^@` below the horizontal . It strikes the ground 3 s later. From what height was the ball thrown? How far from the base of the building does the ball strike the ground?

Explanation

u = 8 m/s, `theta=20^(@)`, t = 35
a) Horizontal distance `=(u costheta)t=8cos`
`" "20^(@) times 3=8 times 0.9397 times 3=22.6m`
b) Height `h=(u sintheta)t +1/2"gt"^(2)`
`=8sin20^(@) times 3+1/2 times 9.8 times 9`
`=8.208+44.1=52.31` m
c) The ball is thrown from a height of 44.1 m
`h_(1)=(usintheta)t_(1)+1/2"gt"_(1)""^(2)`
`10=(8 sin20^(@))t_(1)+1/2""9.8t_(1)""^(2)`
`" "=2.736t_(1)+4.9t_(1)""^(2)`
`rArr 4.9t_(1)""^(2)+2.736t_(1)-10=0`
`t_(1)=(-2.736 pm sqrt((2.736)^(2)+4 times 4.9 times 10))/(2 times 4.9) rArr t_(1)`
`" "=(-2.736 pm sqrt(203.48))/(9.8)`
`t_(1)=(-2.736+14.265)/(9.8)=(11.5288)/(9.8)`
`" "=1.176 =1.18` sec