A ball is thrown straight up and rises to a maximum height of 20 m above the ground. At what height is the speed of the ball equal to half of its initial value? Assume that the ball starts at a height of 2.1 m above the ground.
When the speed is 1/2 of the initial value, 1/4 of the initial kinetic energy will be lost to potential energy, and the ball will be at 3/4 of its maximum distance above where it was thrown from. It rises a total of 17.9 m. 3/4 of that is 14.43 m. Add that to 2.1 m for the answer.
To find out at what height the speed of the ball is equal to half of its initial value, we need to consider the motion of the ball.
The ball is thrown straight up and then falls back down due to the force of gravity. At the highest point in its trajectory, the speed of the ball is momentarily zero. This is because the ball reaches its maximum height and starts to descend.
To find the height at which the speed is equal to half its initial value, we need to find the speed of the ball at its peak and then find the height corresponding to half of that speed.
Step 1: Finding the speed at the peak
We can use the principle of conservation of energy to find the speed of the ball at the peak. At its highest point, the ball has reached its maximum potential energy, which is converted from its initial kinetic energy. So, we can equate the initial kinetic energy with the potential energy at the peak.
Initial Kinetic Energy = Potential Energy at the peak
(1/2)mv^2 = mgh
Here, m is the mass of the ball, v is the speed of the ball at the peak, g is the acceleration due to gravity, and h is the maximum height of the ball.
Since mass (m) cancels out on both sides of the equation, we can simplify it to:
(1/2)v^2 = gh
Solving for v, we get:
v = sqrt(2gh)
Step 2: Finding the height at half the speed
Now that we have the speed at the peak, we can find the height at which the speed is equal to half of that. Let's call this height H.
Using the equation of motion, we can relate the speed, acceleration, and height:
v^2 = u^2 + 2as
Here, u is the initial velocity (which is the speed at the peak), a is the acceleration due to gravity (-9.8 m/s^2), s is the displacement (which is H - 2.1 m), and v is the speed at H (which is half of the initial speed).
Substituting the values and rearranging the equation, we get:
(1/2)v^2 - u^2 = -2as
(1/2)(v^2 - u^2) = -2as
(1/2)(v^2 - sqrt(2gh))^2 = -2(-9.8)(H - 2.1)
Simplifying further, we obtain:
v^2 - sqrt(2gh))^2 = 19.6(H - 2.1)
Now, we substitute the calculated value of v (speed at the peak) and solve for H:
(sqrt(2gh))^2 - sqrt(2gh))^2 = 19.6(H - 2.1)
Simplifying, we get:
0 = 19.6(H - 2.1)
H - 2.1 = 0
H = 2.1 m
Therefore, the height at which the speed of the ball is equal to half of its initial value is 2.1 meters above the ground.