From the top of a tower 156.8 m high a projectile is thrown up with a velocity of 39.2 ms^–1, making an angle 30° with horizontal direction. Find the distance from the foot of tower where it strikes the ground and the time taken by it to do so.

Height of tower OB = 156.8 m; u = 39.2 m/s; θ = 30° component of velocity along OX

= ucosθ = 39.2cos30° = 33.947 ms⁻¹

Component of velocity along OY

= usinθ = 39.2sin30° = 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,

y₀ = 0, y = 156.8 m,
uᵧ = −usin30° = −19.6 m/s,
aᵧ = 9.8 m/s², t = t

As,

y = y₀ + uᵧt + ½aᵧt²

∴ 156.8 = 0 + (−19.6)t + ½ × 9.8 × t²

or 156.8 = −19.6t + 4.9t²

or 4.9t² − 19.6t − 156.8 = 0

or t² − 4t − 32 = 0

or t² − 8t + 4(t − 8) = 0

or t(t − 8) + 4(t − 8) = 0

or (t + 4)(t − 8) = 0

or t = −4 or 8

As t = −4s is not possible, therefore t = 8s.

Distance from the foot of tower where it strikes the ground is

BD = ucos30° × t = 30.947 × 8 = 271.57 m