A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 4 m/s at the floor. Next, the pebble is thrown down with an initial speed of 3 m/s from the same height. What is its speed at the floor? 2 points6 m/s8 m/s5 m/s7 m/s4 m/s
Question
A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 4 m/s at the floor. Next, the pebble is thrown down with an initial speed of 3 m/s from the same height. What is its speed at the floor? 2 points6 m/s8 m/s5 m/s7 m/s4 m/s
Solution
The speed of the pebble when it hits the floor can be calculated using the equation for the final velocity of an object in free fall:
v_f = sqrt(v_i^2 + 2gh)
where:
- v_f is the final velocity (the speed of the pebble when it hits the floor)
- v_i is the initial velocity (the speed of the pebble when it is released)
- g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
- h is the height from which the pebble is dropped
In the first scenario, the pebble is released from rest, so v_i = 0, and we know that v_f = 4 m/s. We can use these values to solve for h:
4 m/s = sqrt(0 + 29.8 m/s^2h) 16 m^2/s^2 = 29.8 m/s^2h h = 16 m^2/s^2 / (2*9.8 m/s^2) = 0.816 m
In the second scenario, the pebble is thrown down with an initial speed of 3 m/s from the same height. We can use the same equation to solve for the final velocity, but this time we know that v_i = 3 m/s and h = 0.816 m:
v_f = sqrt((3 m/s)^2 + 29.8 m/s^20.816 m) v_f = sqrt(9 m^2/s^2 + 15.9744 m^2/s^2) v_f = sqrt(24.9744 m^2/s^2) v_f = 4.997 m/s
So, the speed of the pebble when it hits the floor in the second scenario is approximately 5 m/s. Therefore, the correct answer is 5 m/s.
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