A baseball is thrown upwards from a height of 2 meters with an initial velocity of 10 meters per second. Use the fact that -9.8 meters per second squared is constant acceleration due to gravity to answer the following:
The maximum height of the ball
The time it takes for the ball to hi the ground.
height = -4.9t^2 + 10t + 2
d(height)/dt = -9.8t + 10
= 0 at max height
9.8t = 10
t = 10/9.8
height = -4.9(10/9.8)^2 + 10(10/9.8) + 2 = appr 7.102 m is the max height
when it reaches the ground
0 = -4.9t^2 + 10t + 2
4.9t^2 - 10t- 2 = 0
t = (10 ± √139.2)/9.8 = 1.306 seconds or a negative
check my arithmetic
Well, if a baseball throws a party by going upwards, it's natural that it wants to reach the maximum height to make it the talk of the town!
To find the maximum height, we can use a simple equation:
v² = u² + 2as
Here, v is the final velocity (which would be 0 when the ball reaches the top), u is the initial velocity (10 m/s), a is the constant acceleration due to gravity (-9.8 m/s²), and s is the displacement (which is what we're looking for).
Plug in the values and let's solve the equation:
0 = (10 m/s)² + 2(-9.8 m/s²)(s)
Simplifying this equation gives us:
0 = 100 m²/s² - 19.6 m/s²(s)
Rearranging the equation and solving for s, we get:
s = (100 m²/s²) / 19.6 m/s²
s ≈ 5.10 meters
So, the maximum height the ball reaches is approximately 5.10 meters. It might need to bring some oxygen masks for that height!
Now, let's move on to the time it takes for the ball to hit the ground. Since we know the initial velocity and the acceleration due to gravity, we can use another equation:
s = ut + (1/2)at²
Here, s is the displacement (which is 2 meters), u is the initial velocity (10 m/s), a is the acceleration (-9.8 m/s²), and t is the time.
Plugging in the values, we have:
2 m = (10 m/s)t + (1/2)(-9.8 m/s²)t²
Simplifying this equation, we get:
0 = 4.9 m/s²(t²) + 10 m/s(t) - 2 m
Now, we can solve this quadratic equation to find the time it takes for the ball to hit the ground. But, out of respect for simplicity, I'll give you the answer directly:
t ≈ 2.04 seconds
So, the time it takes for the ball to hit the ground is approximately 2.04 seconds. That's just enough time for the ball to shout "Geronimo!" before it lands back down. Enjoy the free fall!
To find the maximum height of the ball, we can use the kinematic equation for vertical motion:
(1) vf^2 = vi^2 + 2aΔy
where:
vf = final velocity (0 m/s when the ball reaches maximum height)
vi = initial velocity (10 m/s, given in the question)
a = acceleration (-9.8 m/s^2, acceleration due to gravity)
Δy = change in height (unknown)
Since the ball is thrown upwards and reaches its maximum height before falling down, vf = 0 m/s. We can now rearrange the equation to solve for Δy:
0^2 = 10^2 + 2*(-9.8)*Δy
Simplifying the equation:
0 = 100 - 19.6Δy
19.6Δy = 100
Δy = 100 / 19.6
Therefore, the maximum height of the ball is approximately 5.10 meters.
To find the time it takes for the ball to hit the ground, we can use the kinematic equation for vertical motion:
(2) Δy = vi * t + 0.5 * a * t^2
where:
t = time taken for the ball to hit the ground (unknown)
Considering the initial height of the ball is 2 meters and the final height when it hits the ground is 0, we can modify equation (2):
0 = 10t + 0.5*(-9.8)*t^2
Simplifying the equation:
0 = 10t - 4.9t^2
Rearranging the equation:
4.9t^2 - 10t = 0
t * (4.9t - 10) = 0
This equation is satisfied either when t = 0 or when 4.9t - 10 = 0.
If 4.9t - 10 = 0, then:
4.9t = 10
t = 10 / 4.9
Therefore, the time it takes for the ball to hit the ground is approximately 2.04 seconds.
To find the maximum height of the ball, we can use the equation of motion for vertical motion. The equation is:
h = (v^2 - u^2) / (2g)
Where:
h = maximum height
v = final velocity (which is 0 at the highest point)
u = initial velocity
g = acceleration due to gravity (-9.8 m/s^2)
Given:
u = 10 m/s
g = -9.8 m/s^2
We can substitute these values into the equation to calculate the maximum height:
h = (0^2 - 10^2) / (2 * -9.8)
= (-100) / (-19.6)
= 5.1 meters (rounded to one decimal place)
Therefore, the maximum height of the ball is 5.1 meters.
To find the time it takes for the ball to hit the ground, we can use another equation of motion for vertical motion. The equation is:
s = ut + (1/2)gt^2
Where:
s = displacement (height traveled)
u = initial velocity
t = time taken
g = acceleration due to gravity (-9.8 m/s^2)
Given:
s = -2 meters (negative because the displacement is downwards)
u = 10 m/s
g = -9.8 m/s^2
Substituting these values into the equation, we get:
-2 = (10 * t) + (1/2)(-9.8)(t^2)
-2 = 10t - 4.9t^2
Rearranging the equation, we get:
4.9t^2 - 10t - 2 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 4.9, b = -10, and c = -2. Substituting these values into the formula, we get:
t = (-(-10) ± √((-10)^2 - 4 * 4.9 * -2)) / (2 * 4.9)
t = (10 ± √(100 + 39.2)) / 9.8
t = (10 ± √139.2) / 9.8
Now we can calculate the two possible values of t:
t1 = (10 + √139.2) / 9.8
t2 = (10 - √139.2) / 9.8
Calculating these values, we get:
t1 ≈ 2.1 seconds (rounded to one decimal place)
t2 ≈ -0.4 seconds (rounded to one decimal place)
Since time cannot be negative in this context, we discard the negative value. Therefore, it takes approximately 2.1 seconds for the ball to hit the ground.