A pebble of mass 200g is thrown vertically upward such that it returns to the point of projection after 6s. What is the velocity of the pebble
Question
A pebble of mass 200g is thrown vertically upward such that it returns to the point of projection after 6s. What is the velocity of the pebble
Solution
To solve this problem, we need to use the equations of motion.
Step 1: Identify the given information
- The total time taken (t) = 6 seconds
- The mass of the pebble (m) = 200g = 0.2 kg (since 1 kg = 1000 g)
- The acceleration due to gravity (g) = 9.8 m/s²
Step 2: Understand the problem The pebble is thrown upwards and comes back to the starting point. This means the total time taken includes the time taken to go up and come down. So, the time taken to reach the maximum height is half the total time, which is 6/2 = 3 seconds.
Step 3: Use the equation of motion We can use the second equation of motion, which is v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
Since the pebble comes back to the starting point, the final velocity (v) is 0. We can rearrange the equation to solve for the initial velocity (u):
u = v - gt u = 0 - (-9.8 * 3) u = 29.4 m/s
So, the initial velocity of the pebble, or the velocity at which it was thrown upwards, is 29.4 m/s.
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