Question

A ball is thrown upward from the top of a building at an angle of \(30^\circ\) to the horizontal and with an initial speed of \(20\ \text{m s}^{-1}\). If the ball strikes the ground after \(3\) s, then the height of the building is
\[ (\text{acceleration due to gravity }=10\ \text{m s}^{-2}) \]

• A. \(10\ \text{m}\)
• B. \(15\ \text{m}\)
• C. \(20\ \text{m}\)
• D. \(25\ \text{m}\)

Hint:

In projectile motion, always resolve the initial velocity into horizontal and vertical components first: \[ u_x=u\cos\theta,\qquad u_y=u\sin\theta. \] Then apply the vertical motion equation \[ s=u_y t-\frac{1}{2}gt^2 \] to determine heights and vertical displacements.

Answer 1

The Correct Option is B

Step 1: Resolve vertical motion.
Initial vertical velocity: \( u_y = 20\sin 30^\circ = 10\text{ m/s} \) (upward). Taking downward as positive, displacement after 3 s equals height \( H \).

Step 2: Apply \( s = -u_y t + \frac{1}{2}gt^2 \).
\( H = -10(3) + \frac{1}{2}(10)(9) = -30 + 45 = 15\text{ m} \)

\[ \boxed{H = 15\text{ m}} \]