If a ball is dropped from a 100 m cliff, it accelerates at a constant rate of 9.8 m/s2. Assuming its initial velocity is 0 m/s, its initial position is 0 m and its final position is 100 m away from the initial position, what is the velocity of the ball when it hits the ground?
Question
If a ball is dropped from a 100 m cliff, it accelerates at a constant rate of 9.8 m/s2. Assuming its initial velocity is 0 m/s, its initial position is 0 m and its final position is 100 m away from the initial position, what is the velocity of the ball when it hits the ground?
Solution
To find the final velocity of the ball when it hits the ground, we can use the equation of motion:
v² = u² + 2as
where: v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
Given in the problem: u = 0 m/s (initial velocity), a = 9.8 m/s² (acceleration due to gravity), and s = 100 m (height of the cliff).
Substituting these values into the equation, we get:
v² = 0 + 29.8100 v² = 1960 v = sqrt(1960) v = 44.27 m/s
So, the velocity of the ball when it hits the ground is approximately 44.27 m/s.
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Every word problem, there should be a conclusion. therefore add a conclusion to the following answer. "(a) The velocity of the ball at its highest point is 0 m/s. This is because at the highest point, the ball stops moving upwards and is about to start falling downwards. (b) The velocity of the ball 1 second before it reaches its highest point depends on the initial speed and the acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s². So, if we denote the initial speed as v0, then the velocity 1 second before the highest point is v0 - 9.8 m/s. (c) The change in velocity during this 1-second interval is the final velocity minus the initial velocity, which is (v0 - 9.8 m/s) - v0 = -9.8 m/s. (d) The velocity of the ball 1 second after it reaches its highest point is -9.8 m/s. This is because the ball has started to fall and is accelerating downwards due to gravity. (e) The change in velocity during this 1-second interval is the final velocity minus the initial velocity, which is -9.8 m/s - 0 = -9.8 m/s. (f) The change in velocity during the 2-second interval is the final velocity after 2 seconds minus the initial velocity, which is (-9.8 m/s) - (v0 - 9.8 m/s) = -19.6 m/s. (g) The acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity is -9.8 m/s². This is the acceleration due to gravity, and it is always acting on the ball, regardless of its velocity."
A ball is dropped from a helicopter 100m from the ground. As the ball falls, the magnitude of it's associated velocity isQuestion 14Answera.increasing at a rate of 9.81 m/s each secondb.a constantc.decreasing at a rate of 9.81 m/s each secondd.equal to 9.81 m/se.is zero
What is the velocity of the ball when it hits the ground?
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