A ball is thrown straight upward. At 5.00 m above its launch point, the ball’s speed is one-half its launch speed. What maximum height above its launch point does the ball attain?
V^2 = Vo^2 + 2g*h
(Vo/2)^2 = Vo^2 + 2g*h
Vo^2/4 = Vo^2 - 19.6*5
Vo^2/4 = Vo^2 - 98
Multiply both sides by 4:
Vo^2 = 4Vo^2 - 392
4Vo^2 - Vo^2 = 392
3Vo^2 = 392
Vo^2 = 130.7
Vo = 11.43 m/s.
hmax = (V^2-Vo^2)/2g
h max = (0-11.43^2)/-19.6 = 6.67 m.
Why did the ball bring math into this? Can't it just enjoy its time in the air without all these complicated calculations? Anyway, to answer your question, let's say the launch speed is "X." So, when the ball is at a height of 5.00 m, its speed will be X/2. Now, let's hope the ball had good aim and reached its maximum height because I don't want it getting any ideas about becoming an astronaut. In order to find the maximum height, we need to know the initial velocity, acceleration due to gravity, and the time it takes for the ball to reach its highest point. If we have all those, we can just plug them into a formula and voila! We'll have our answer. So, don't worry, we'll get there.
To find the maximum height above its launch point that the ball attains, we need to use the concept of conservation of energy.
Let's assume:
Initial launch speed = V₀
Final speed at a height of 5.00 m = V
According to the problem, the ball's speed at 5.00 m above its launch point is one-half its launch speed. Therefore, we have:
V = (1/2) * V₀
At maximum height, the ball's final speed is zero because it momentarily comes to a stop before falling back down. Therefore, we can set the final speed (V) to zero and solve for the maximum height (H).
Kinetic energy at launch = Potential energy at maximum height
(1/2) * m * (V₀)^2 = m * g * H
Where:
m = mass of the ball
g = acceleration due to gravity
Since the mass of the ball cancels out, we can remove it from the equation. Rearranging the equation, we get:
(1/2) * (V₀)^2 = g * H
Now, we can substitute the value of V in terms of V₀:
(1/2) * (1/4) * (V₀)^2 = g * H
Simplifying further,
(1/8) * (V₀)^2 = g * H
Finally, solving for H, we get:
H = (1/8) * (V₀)^2 / g
Therefore, the maximum height above its launch point that the ball attains is (1/8) * (V₀)^2 / g.
To find the maximum height above its launch point that the ball attains, we can use the equations of motion.
Let's assume:
- Initial velocity of the ball when it is thrown upward is 'v₀'
- The maximum height the ball reaches above its launch point is 'h'
- The speed of the ball when it is at 5.00 m above its launch point is 'v'
The first thing we need to determine is the relationship between the speed of the ball at 5.00 m above its launch point and its initial velocity.
We know that at the highest point of the ball's trajectory, its velocity will be zero. So, when the ball is at a height of 5.00 m above its launch point, its velocity will be half its launch speed.
Using the equation for velocity of an object in free fall:
v = v₀ - gt
where 'g' is the acceleration due to gravity (approximately 9.8 m/s²), we can substitute the given values:
v/2 = v₀ - 9.8 × t
Since the ball is at 5.00 m above its launch point, we can determine the time it takes to reach that height. Using the equation for displacement:
s = v₀t - 0.5gt²
Substituting the values:
5.00 = v₀t - 0.5 × 9.8 × t²
Now, let's solve these two equations simultaneously to find the initial velocity and the time taken to reach a height of 5.00 m.
From the first equation, we can solve for 't' in terms of 'v' and 'v₀':
t = (v₀ - (v/2)) / 9.8
Substituting this value of 't' into the second equation:
5.00 = v₀ × ((v₀ - (v/2)) / 9.8) - 0.5 × 9.8 × ((v₀ - (v/2)) / 9.8)²
Simplifying this equation will allow us to solve for 'v₀':
5.00 = v₀ × ((v₀ - (v/2)) / 9.8) - 0.5 × ((v₀ - (v/2)) / 9.8)²
Now, let's solve this equation for 'v₀'. We can use a numerical method like trial and error or solve it graphically.