Question

From the top of a tower 156.8m high, a projectile is thrown up with a velocity of 39.2m/s, making an angle 30 degree with the horizontal direction. find the distance from the foot of tower where it strikes the groung and the time taken by it to do so ?

Answer 1

equation of motion for the projectile 
y  = x  tanФ  -  g x² Sec² Ф / (2 u²)
=>  y = x/√3   -  9.8 (4/3) x²/ (2 * 39.2²)             as  Ф  = 30°
=> y = x/√3  - x² /235.2

At  t=0,   x =0  and y = 0 
We have to find x when y = - 156.8 m

so:   x²/235.2 - 156.8 - x/√3  = 0
 =>  √3 x²  - 235.2 x - 235.2 * 156.8√3 = 0

   x = [ 235.2 + - √(235.2² + 4 * 235.2 * 156.8 * 3) ] / (2√3)
     = [ 235.2 + - 235.2 *√(1+8) ]/(2 √3)
     =   2 * 235.2 /√3  meters   or   ignoring negative value.
This is the distance from the foot...

Time taken to hit the ground =  x /(u cosФ)
  = [ 2 * 235.2/√3 ] / (39.2 * √3 / 2)  sec
  = 8 sec