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lakkis [162]
1 year ago
7

Amir pitches a baseball at an initial height of 6 feet with a velocity of 73 feet per second. this can be represented by the fun

ction h(t) = −16t2 73t 6. if the batter misses, about how long does it take the ball to hit the ground? 4.64 seconds 2.94 seconds 2.28 seconds 0.08 seconds
Physics
1 answer:
Romashka [77]1 year ago
8 0

The values of t are <u>4.643 second</u> for the function H(t)=-16t^2+73t+6

What is batter misses?

An out in baseball happens when the umpire declares a batter or baserunner out. A hitter or runner who is out is no longer able to score runs and must go back to the dugout until their subsequent turn at bat. The batting team's turn is over after three outs are recorded in a half-inning.

In order to signal an out, umpires typically make a fist with one hand and then flex that arm, either upward on pop flies or forward on regular plays at first base. To indicate a called strikeout, home plate umpires frequently use a "punch-out" action.When a batter is struck by a pitched ball without making a swing at it, it is referred to as a hit-by-pitch. He consequently gets first base.

We have been given that

s = 6 feet

v = 73 feet per second

Substituting these values in the formula H(t)=-16t^2+vt+s

H(t)=-16t^2+73t+6

When the ball hits the ground, the height becomes zero. Thus, H(t)=0

-16t^2+73t+6=0

We solve the equation using quadratic formula x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

Substituting the values a=-16, b= 73, c=6

t_{1,2}=\frac{-73 \pm \sqrt{(73)^2-4(-16)(6)}}{2(-16)}\\\Rightarrow t_{1,2}=\frac{-73 \pm \sqrt{5713}}{2(-16)}\\\Rightarrow t_{1,2}=-0.081, 4.643

Learn more about the batter misses with the help of the given link:

brainly.com/question/19475098

#SPJ4

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a) 0.0028 rad/s

b) 23.68 m/s^2

c) 0 m/s^2

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

\omega = \frac{\theta}{t}

where

\theta is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

\omega = \frac{v}{r} (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

v=29,960 km/h is the linear speed, converted into m/s,

v=8322 m/s

r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the  centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

a_r=\omega^2 r

where

\omega is the angular speed

r is the radius of the orbit

For the spaceship in the problem, we have

\omega=0.0028 rad/s is the angular speed

r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

a_t=\frac{\Delta v}{\Delta t}

where

\Delta v is the change in the linear speed

\Delta  t is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

v=8322 m/s

Therefore, its linear speed is not changing, so the change in linear speed is zero:

\Delta v=0

And therefore, the tangential acceleration is zero as well:

a_t=\frac{0}{\Delta t}=0 m/s^2

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