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

Initial velocity, u=`8m//s`,
Angle of protection , `theta=20^@`
Time taken to reach the ground , t=3 sec Horizontal component of initial velocity,
`u_x=u.costheta=8cos20^@`
`=8times0.94=7.52m//s`
Vertical component of initial velocity,
`v_y=usintheta=8sin20^@`
`=8times0.342=2.736m//s`
a) From equation of motion , `s=ut+1/2 at^2`
we can write
`h=(usintheta)t+1/2g t^2`
`impliesh=(2.736)3+1/2times9.8times(3)^2`
`impliesh=8.208+4.9 times9`
`implies h=8.208+44.1 or h=52.308m`

b) Horizontal distance travelled,
`s_x=v_xtimest=7.52times3=22.56m`