A ball is thrown straight upward and rises to a maximum height of 14.6 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?
To find the height at which the speed of the ball decreases to one-half of its initial value, we need to understand the motion of the ball and make use of the relevant physics equations.
First, let's assume that the initial velocity of the ball when it is thrown upward is "v0" and its final velocity at the maximum height is zero. The acceleration due to gravity (g) is acting in the downward direction.
The speed of an object can be calculated using the equation v = u + at, where:
- v is the final velocity,
- u is the initial velocity,
- a is the acceleration, and
- t is the time taken.
In this problem, the ball reaches a maximum height when its speed becomes zero. To find the time it takes for the ball to reach this maximum height, we can use the equation v = u + at and set v = 0. Thus, we have 0 = v0 + (-g)t, which can be rearranged as t = v0/g.
Now, we want to find the height at which the speed is reduced to one-half of its initial value. Let's call this height "h". The time taken to reach this height will be denoted as "t1".
To find the time t1, we can rearrange the equation v = u + at and set v = v0/2. This gives us v0/2 = v0 + (-g)t1, which simplifies to -gt1 = v0/2. Solving for t1, we get t1 = v0/(2g).
Next, we need to find the height h using the equation for vertical displacement, which is given by h = u*t1 + (1/2)*a*t1^2.
Since the initial velocity u is v0 and the acceleration a is -g (negative due to the direction of the acceleration), we have h = v0*t1 - (1/2)*g*t1^2.
Finally, we can substitute the value of t1 in terms of v0/g, which we previously found, to get the final expression for h in terms of v0:
h = (v0^2) / (2g)
Using this formula, we can calculate the height at which the speed of the ball decreases to one-half of its initial value, by substituting the appropriate values for v0 and g.
Therefore, the height above its launch point at which the speed of the ball decreases to one-half of its initial value is given by h = (v0^2) / (2g), where v0 is the initial velocity and g is the acceleration due to gravity.