In slow- pitch softball the ball is pitched in an underhand motion. A batter in a softball game is pitched a ball that has initial vertical velocity of 35 feet per second.
a. Write an equation for the height h (in feet) of the ball as a function of the time t (in seconds) after it is pitched.
b. The batter hits the ball when it is 2.5 feet above the ground. How long after the ball is pitched is the ball hit? Round your answer to the nearest tenth of a second.
a. The equation for the height h (in feet) of the ball as a function of time t (in seconds) after it is pitched can be represented by the equation:
h(t) = -16t^2 + 35t
b. To find the time t when the ball is hit, we need to solve the equation h(t) = 2.5. Plugging in this value of h(t) into the equation from part a, we have:
-16t^2 + 35t = 2.5
Now, let's solve the equation:
-16t^2 + 35t - 2.5 = 0
Using the quadratic formula, we get:
t = (-35 ± √(35^2 - 4(-16)(-2.5))) / (2(-16))
Simplifying further:
t = (-35 ± √(1225 - 160)) / (-32)
t = (-35 ± √(1065)) / (-32)
Calculating the values using a calculator:
t ≈ 0.285 seconds (rounded to the nearest tenth)
Therefore, the ball is hit approximately 0.3 seconds after it is pitched.
a. To determine the equation for the height of the ball as a function of time, we need to consider the vertical motion of the ball. In this case, the motion is affected by gravity, and the initial vertical velocity is 35 feet per second. The equation for the height (h) of the ball can be determined using the following equation of motion:
h(t) = h0 + v0t - (1/2)gt^2
Where:
h(t) is the height of the ball at time t
h0 is the initial height of the ball (when t = 0)
v0 is the initial vertical velocity of the ball
g is the acceleration due to gravity (approximately 32.2 feet per second squared)
In this case, the initial height of the ball is not given, so we assume that it is 0:
h(t) = 0 + 35t - (1/2)(32.2)t^2
h(t) = 35t - 16.1t^2
b. To find the time when the ball is hit (h = 2.5 feet), we set h(t) equal to 2.5 and solve for t:
2.5 = 35t - 16.1t^2
To solve this quadratic equation, we can rearrange it to the standard form:
16.1t^2 - 35t + 2.5 = 0
We can then solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where:
a = 16.1
b = -35
c = 2.5
Substituting these values into the quadratic formula, we have:
t = (-(-35) ± √((-35)^2 - 4(16.1)(2.5))) / (2(16.1))
Simplifying further, we get:
t = (35 ± √(1225 - 322)) / 32.2
t ≈ (35 ± √903) / 32.2
Using a calculator, we can find the approximate values of t:
t ≈ (35 + √903) / 32.2 ≈ 1.6 seconds or t ≈ (35 - √903) / 32.2 ≈ 0.053 seconds
Rounding to the nearest tenth of a second, the ball is hit approximately 1.6 seconds after it is pitched.
To answer this question, we need to understand the basic principles of kinematics, specifically projectile motion. The equation for the height of the ball as a function of time can be derived using the equation for vertical motion:
h(t) = h0 + v0 * t + (1/2) * a * t^2
Where:
h(t) is the height of the ball at time t
h0 is the initial height (in this case, we assume it to be 0 since we are measuring the height relative to the ground)
v0 is the initial vertical velocity (given as 35 ft/s)
t is the time in seconds after the ball is pitched
a is the vertical acceleration, which we assume to be -32 ft/s^2 (the acceleration due to gravity)
Now, let's substitute the given values into the equation:
h(t) = 0 + 35t + (1/2) * (-32) * t^2
Simplifying further:
h(t) = 35t - 16t^2
This equation represents the height of the ball as a function of time after it is pitched.
To determine the time when the ball is hit, we set the height h(t) equal to 2.5 feet:
2.5 = 35t - 16t^2
Next, we rearrange the equation to form a quadratic equation:
16t^2 - 35t + 2.5 = 0
We can now solve this quadratic equation for t. Using the quadratic formula:
t = [-(-35) ± sqrt((-35)^2 - 4(16)(2.5)) ] / (2*16)
Simplifying further:
t = [35 ± sqrt(1225 - 160)] / 32
t = [35 ± sqrt(1065)] / 32
Calculating the values:
t ≈ [35 + sqrt(1065)] / 32 ≈ 2.3 seconds (rounded to the nearest tenth)
Therefore, the ball is hit approximately 2.3 seconds after it is pitched.