Question

A body is projected from a point which is at a height of \(5\) m from the ground with a velocity of \(28\ \mathrm{ms^{-1}}\) at an angle of \(30^\circ\) above the horizontal. The horizontal distance travelled by the body when it reaches a point which is at a height of \(2\) m from the ground is \[ (g=10\ \mathrm{ms^{-2}}) \]

• A. \(126\sqrt3\) m
• B. \(21\sqrt3\) m
• C. \(42\sqrt3\) m
• D. \(28\sqrt3\) m

Hint:

For projectile motion, \[ \boxed{ y=y_0+u\sin\theta\,t-\frac12gt^2 } \] is used to find the time, and then \[ \boxed{x=u\cos\theta\,t} \] gives the horizontal distance.

Answer 1

The Correct Option is C

Step 1: Resolve the initial velocity. Initial speed: \[ u=28\text{ ms}^{-1}. \] Horizontal component: \[ u_x = 28\cos30^\circ = 14\sqrt3\text{ ms}^{-1}. \] Vertical component: \[ u_y = 28\sin30^\circ = 14\text{ ms}^{-1}. \] Initial height: \[ y_0=5\text{ m}. \]

Step 2: Find the time when the body is at a height of \(2\) m. Using \[ y = y_0+u_yt-\frac12gt^2, \] we get \[ 2 = 5+14t-5t^2. \] Thus, \[ 5t^2-14t-3=0. \] Solving, \[ t=\frac{14\pm16}{10}. \] Taking the positive value, \[ t=3\text{ s}. \]

Step 3: Find the horizontal distance. Horizontal distance is \[ x=u_xt. \] Hence, \[ x = 14\sqrt3\times3 = 42\sqrt3\text{ m}. \] Therefore, \[ \boxed{42\sqrt3\text{ m}}. \] Thus, \[ \boxed{(C)} \] is the correct answer.