Question:medium

If a pebble is thrown vertically upwards from the top of a tower with velocity \(5\,m/s\), it strikes the ground after \(3\) seconds. With what velocity the pebble strikes the ground? \((g=10\,m/s^2)\)

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For vertical motion, carefully choose sign convention. If downward is positive, upward velocity is negative.
  • \(10\,m/s\)
  • \(20\,m/s\)
  • \(25\,m/s\)
  • \(30\,m/s\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
This is a problem of motion under constant acceleration (gravity). We need to use the kinematic equations for uniformly accelerated motion to find the final velocity.
Step 2: Key Formula or Approach:
The relevant kinematic equation is: \[ v = u + at \] where: - $v$ is the final velocity. - $u$ is the initial velocity. - $a$ is the acceleration. - $t$ is the time. We need to establish a sign convention. Let's take the upward direction as positive and the downward direction as negative.
Step 3: Detailed Explanation:
The given parameters are: - Initial velocity, $u = +5$ m/s (since it's thrown upwards). - Time of flight, $t = 3$ s. - Acceleration due to gravity, $a = -g = -10 \text{ ms}^{-2}$ (since gravity acts downwards). We need to find the final velocity, $v$, when it strikes the ground. Using the kinematic equation: \[ v = u + at \] Substitute the values: \[ v = (+5) + (-10)(3) \] \[ v = 5 - 30 \] \[ v = -25 \text{ m/s} \] The negative sign indicates that the final velocity is in the downward direction, which is expected as the pebble is striking the ground. The question asks for the velocity (speed) with which it strikes, which is the magnitude of the final velocity. \[ \text{Speed} = |v| = |-25| = 25 \text{ m/s} \] Step 4: Final Answer:
The pebble strikes the ground with a velocity of 25 m/s. Therefore, option (C) is correct.
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