Question

A ball is projected horizontally with a velocity of $ 5 \, \text{m/s} $ from the top of a building 19.6 m high. How long will the ball take to hit the ground?

• A. \( \sqrt{2} \, \text{s} \)
• B. \( 2 \, \text{s} \)
• C. \( \sqrt{3} \, \text{s} \)
• D. \( 3 \, \text{s} \)

Hint:

When a ball is projected horizontally, the time it takes to hit the ground depends only on the height of the fall and the acceleration due to gravity. The horizontal velocity does not affect the falling time.

Answer 1

The Correct Option is B

The ball is projected horizontally, so the horizontal velocity does not affect the time taken to hit the ground. 
The time to fall depends only on the vertical motion, which is influenced by gravity. 
The equation for the time taken for an object to fall freely from a height \( h \) is: \[ t = \sqrt{\frac{2h}{g}} \] where: 
- \( h = 19.6 \, \text{m} \) is the height of the building, 
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Substituting the values into the equation: \[ t = \sqrt{\frac{2 \times 19.6}{9.8}} = \sqrt{\frac{39.2}{9.8}} = \sqrt{4} = 2 \, \text{seconds} \] 
Thus, the time taken for the ball to hit the ground is \( 2 \, \text{seconds} \). Therefore, the correct answer is: \[ \text{(2) } 2 \, \text{s} \]

Answer 2

To determine how long the ball takes to hit the ground, we can analyze its vertical motion. The ball is projected horizontally, so its initial vertical velocity (\(v_{0y}\)) is \(0\) m/s. We use the equation of motion for vertical displacement (\(y\)) given by:

\(y = v_{0y}t + \frac{1}{2}gt^2\)

where \(y = 19.6 \, \text{m}\) (the height of the building) and \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity). Substituting into the equation:

\(19.6 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2\)

\(19.6 = 4.9t^2\)

Solving for \(t^2\), we get:

\(t^2 = \frac{19.6}{4.9}\)

\(t^2 = 4\)

Taking the square root of both sides gives:

\(t = \sqrt{4}\)

\(t = 2 \, \text{s}\)

Hence, the ball takes \(2 \, \text{s}\) to hit the ground.