Hakeem throws a 11.0 g ball straight down from a height of 2.0 m. The ball strikes the floor at a speed of 7.8 m/s. What was the initial speed of the ball? Assume that air resistance is negligible.
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First of all, Henry cannot be right because the ball was "thrown" at the ground. His formula is only valid if it was free fall. However, the ball gained acceleration from the throw. You have to use energy formulas. I have a similar question on my worksheet, but I don't have a mass so I think the my problem is wrong. You should use energy problems in your case, but I'm too lazy to do all the work. Here are the formulas you need.
E=mgh
E=(1/2)mv^2
Also, remember conservation of energy.
To find the initial speed of the ball, we can use the equations of motion. We know the height from which the ball is dropped (2.0 m), and the final speed when it strikes the floor (7.8 m/s). We need to find the initial speed.
The equation that relates initial velocity (u), final velocity (v), acceleration (a), and displacement (s) is:
v^2 = u^2 + 2as
In this case, the initial velocity is what we want to find, the final velocity is 7.8 m/s, the acceleration is the acceleration due to gravity (approximately 9.81 m/s^2), and the displacement is -2.0 m (negative because the ball is moving downwards).
Plugging the values into the equation, we have:
(7.8 m/s)^2 = u^2 + 2 * (-9.81 m/s^2) * (-2.0 m)
Solving for u:
60.84 m^2/s^2 = u^2 + 39.24 m^2/s^2
u^2 = 60.84 m^2/s^2 - 39.24 m^2/s^2
u^2 = 21.60 m^2/s^2
Taking the square root of both sides:
u = √(21.60 m^2/s^2)
u ≈ 4.65 m/s
Therefore, the initial speed of the ball was approximately 4.65 m/s.
Vf^2 = Vo^2 + 2g*h,
Vo^2 = Vf^2 - 2g*h,
Vo^2 = (7.8)^2 - 19.6*2 = 21.64,
Vo = 4.65 m/s.