Question

A ball is projected horizontally from the top of a tower with a velocity of \( 10 \text{ m/s} \). If it hits the ground \( 2 \text{ seconds} \) later, what is the height of the tower? (Take \( g = 10 \text{ m/s}^2 \))

• A. \( 20 \text{ m} \)
• B. \( 10 \text{ m} \)
• C. \( 40 \text{ m} \)
• D. \( 25 \text{ m} \)

Hint:

In horizontal projection problems: \[ \text{Horizontal motion} \rightarrow \text{constant velocity} \] \[ \text{Vertical motion} \rightarrow \text{free fall under gravity} \] The height depends only on vertical motion and time of flight.

Answer 1

The Correct Option is A

Concept: In horizontal projection, the horizontal and vertical motions are independent of each other. The vertical motion is simply a case of free fall under gravity. The vertical displacement is given by: \[ h = ut + \frac{1}{2}gt^2 \] Since the ball is projected horizontally, its initial vertical velocity is zero: \[ u=0 \]

Step 1: Write the equation for vertical displacement.
Using the formula: \[ h = ut + \frac{1}{2}gt^2 \] Substitute: \[ u=0, \qquad g=10\text{ m/s}^2, \qquad t=2\text{ s} \] Thus: \[ h = 0 + \frac{1}{2}(10)(2)^2 \]

Step 2: Simplify the numerical expression.
\[ h = 5 \times 4 \] \[ h = 20\text{ m} \] Therefore, the height of the tower is: \[ \boxed{20\text{ m}} \]