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 m s\(^{-1}\). If the ball strikes the ground after 3 s, then the height of the building is (acceleration due to gravity = 10 ms\(^{-2}\))

• A. 10 m
• B. 15 m
• C. 20 m
• D. 25 m

Hint:

Resolve initial velocity into horizontal (\(u_x = u\cos\theta\)) and vertical (\(u_y = u\sin\theta\)) components.
Apply equations of motion separately for horizontal and vertical motion.
For vertical motion under gravity: \(s_y = u_y t + \frac{1}{2} a_y t^2\), where \(a_y = -g\) if upward is positive.
Be careful with signs for displacement, velocity, and acceleration.

Answer 1

The Correct Option is B

Let the height of the building be H. The ball is thrown from this height. Initial speed \(u = 20 \text{ m s}^{-1}\). Angle of projection \(\theta = 30^\circ\) to the horizontal. Acceleration due to gravity \(g = 10 \text{ ms}^{-2}\) (acting downwards). Time of flight until it strikes the ground \(t = 3\) s. Consider the vertical motion of the ball. Take the upward direction as positive and the point of projection as the origin for vertical displacement. Initial vertical velocity \(u_y = u \sin\theta = 20 \sin(30^\circ) = 20 \times (1/2) = 10 \text{ m s}^{-1}\). Vertical acceleration \(a_y = -g = -10 \text{ ms}^{-2}\). When the ball strikes the ground, its vertical displacement from the top of the building is \(-H\) (since ground is below the point of projection). Let \(s_y\) be the vertical displacement. \(s_y = -H\). Using the equation of motion \(s_y = u_y t + \frac{1}{2} a_y t^2\): \( -H = (10 \text{ ms}^{-1})(3 \text{ s}) + \frac{1}{2}(-10 \text{ ms}^{-2})(3 \text{ s})^2 \) \( -H = 30 + \frac{1}{2}(-10)(9) \) \( -H = 30 - 5 \times 9 \) \( -H = 30 - 45 \) \( -H = -15 \) \( H = 15 \text{ m} \). The height of the building is 15 m. This matches option (b). \[ \boxed{15 \text{ m}} \]