A person standing at the top of a hemispherical rock of radius R = 5.00 m kicks a ball (initially at rest on the top of the rock) to give it horizontal velocity vi.
(a) What must be its minimum initial speed if the ball is never to hit the rock after it is kicked?
(b) With this initial speed, how far from the base of the rock does the ball hit the ground?
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To solve this problem, let's break it down into two parts:
(a) To find the minimum initial speed of the ball, we need to determine the lowest speed at which it will clear the hemispherical rock without hitting it.
To do this, we can use the principle of conservation of mechanical energy, which states that the total mechanical energy (TME) of a system remains constant if no external forces are acting on it.
Initially, the ball is at the top of the rock, which is a height of R from its center. At this point, the ball has only potential energy given by mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
When the ball hits the ground, it has zero potential energy and only kinetic energy given by (1/2)mv^2, where v is its velocity.
Since mechanical energy is conserved, we can equate the initial potential energy to the final kinetic energy:
mgh = (1/2)mv^2
Simplifying and canceling the mass, we get:
gh = (1/2)v^2
Substituting g = 9.8 m/s^2 (acceleration due to gravity) and h = R = 5.00 m, we have:
(9.8 m/s^2) * (5.00 m) = (1/2)v^2
Solving for v, we find:
v = √(2 * 9.8 m/s^2 * 5.00 m) = √(98 m^2/s^2) = 9.90 m/s
So, the minimum initial speed of the ball is 9.90 m/s.
(b) To find how far from the base of the rock the ball hits the ground, we can use the horizontally launched projectile motion equations.
The horizontal distance covered by a projectile launched with an initial horizontal velocity vi and time of flight t is given by:
distance = vi * t
In this case, the time of flight (the time it takes for the ball to hit the ground) can be found using the vertical motion equations.
The vertical distance (h') from the top of the rock to the ground is R (radius of the rock), which is 5.00 m.
We can use the equation for vertical motion to find the time of flight:
h' = (1/2)gt^2
Plugging in h' = R = 5.00 m and g = 9.8 m/s^2, we have:
5.00 m = (1/2)(9.8 m/s^2)t^2
Simplifying and solving for t, we get:
t = √(2 * 5.00 m / 9.8 m/s^2) = √(1.02 s^2) = 1.01 s
Now, we can find the horizontal distance covered by the ball:
distance = vi * t
Using vi = 9.90 m/s (from part a) and t = 1.01 s, we have:
distance = 9.90 m/s * 1.01 s = 9.999 m
Rounding to the appropriate number of significant figures, the ball hits the ground at a distance of approximately 10.0 m from the base of the rock.
Therefore, the answer to part (b) is 10.0 m.