Question

A ball is thrown from the top of a building of height 25 m with an initial velocity of 15 m/sec. If the height of the ball h from ground at any point of time t is given by \(h=25+10t-3t^2\). The time taken by the ball to reach the ground is

• A. 15 sec
• B. 10 sec
• C. 5 sec
• D. \(\frac{5}{3}\ \text{sec}\)

Answer 1

The Correct Option is C

We are given the height function as: \[ h = 25 + 10t - 3t^2 \] To find when the ball reaches the ground, set \( h = 0 \): \[ 0 = 25 + 10t - 3t^2 \] Rewriting: \[ 3t^2 - 10t - 25 = 0 \] Using the quadratic formula: \[ t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot (-25)}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 + 300}}{6} = \frac{10 \pm \sqrt{400}}{6} \] \[ t = \frac{10 \pm 20}{6} \] So the two roots are: \[ t = \frac{10 + 20}{6} = \frac{30}{6} = 5 \quad \text{and} \quad t = \frac{10 - 20}{6} = \frac{-10}{6} = -\frac{5}{3} \] Since time cannot be negative, the valid solution is: \[ t = 5 \text{ seconds} \]

The correct option is (C): \(5\ sec\)