Question

A ball is thrown vertically upward with a velocity of 20 m/s from the top of a multistory building. The height of the point from where the ball is thrown is 25 m above the ground. Find the time taken before the ball hits the ground. (Take \( g = 10 \, \text{m/s}^2 \)).

• A. 3 s
• B. 4 s
• C. 5 s
• D. 6 s

Hint:

Key Fact: When solving projectile motion problems, account for the initial height and use the sign convention carefully (upward velocity positive, downward acceleration negative).

Answer 1

The Correct Option is C

Step 1: Set Up the Equation of Motion
Use the kinematic equation for displacement: \[ s = ut + \frac{1}{2}at^2 \] Here, \( u = 20 \, \text{m/s} \) (upward), \( a = -g = -10 \, \text{m/s}^2 \) (downward), initial height \( s_0 = 25 \, \text{m} \), and final displacement \( s = -25 \, \text{m} \) (when the ball hits the ground). The total displacement relative to the starting point: \[ -25 = 20t + \frac{1}{2}(-10)t^2 \]

Step 2: Simplify the Equation
\[ -25 = 20t - 5t^2 \implies 5t^2 - 20t - 25 = 0 \implies t^2 - 4t - 5 = 0 \]

Step 3: Solve the Quadratic Equation
Solve \( t^2 - 4t - 5 = 0 \) using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), \( c = -5 \): \[ t = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2} \] \[ t = 5 \quad \text{or} \quad t = -1 \] Since time cannot be negative, \( t = 5 \, \text{s} \).

Step 4: Final Answer
The time taken for the ball to hit the ground is 5 seconds.