For the top of a tower ` 156.8 m` high, a projectile is thrown up with a velcity of `39.2 ms^(-1)`, makingan angle `30%(@)` with borizontal direction. Find the distance from the foot of tower wher it strikes the ground and the time taken byit do so.

Explanation

If Fig. 2(d). 26, height tower
`OB =156.8 m, u= 39.2 m//s , theta =30^(@)`
Component of velocity along `OX`
`=u cos theta =30^(@).2 cos 30 ^(@) =33.947 m^(-10`
.
Component of velocity along `OY`
` = u sin theta =39 .2 sin 30^(@) =19. 6 ms^(-1)`
Let (t) be the total time of flight (i.e time in going from ` O to D). Consider the vertical downward direction `OB` as the positive direction of y-axis. Taking motion of a projectile from `O to D` along Y-axis, we have
`y_(0) =0, y= 156.8 m`,
`u_(y) =- sin 30^(@) =- 19.6 m//s`,
`a_(y) =9.8 m//s^(2) , t=t`
As, y=y_(0) +u_(y) t+ 1/2 _(Y) T^(2)` ltBRgt :. 156. 8 =0 + (-19.6) t+ 1/2 xx 9.8 xx t^(2)`
or 156.8 =- 19.6 + 4.9 t^(2)`
or `4.9 t^(2) -19.6 t156.8=0`
or `t^(2) -4 t-32 =0`
or` t^(2) -8 t+ 4 t -32 =0`
or ` t (t-8) + 4 (t -8) =0`
or `(t-4) (t-8) =0 or t=-4 or 8`.
As `t=- 4 s is not possible therefore `t=8 s`
Distance from the foot of tower wher it strikes the ground is `
`BD =u cos 30^(@) xx t=33. 947 xx =271.57 m`.