An object start from rest with a constant acceleration of 8.00 m/s^2 along straight line.

baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the

Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0

3 years ago

Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. the tide measures 5.15 ft at

RSB [31]

<span>f(x) = 5.05*sin(x*pi/12) + 5.15

First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.

So our function at this point looks like f(x) = 5.05*sin(x*pi/12) But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
f(x) = 5.05*sin(x*pi/12) + 5.15

The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
f(x) = 5.05*sin(x*pi/12 + C) + 5.15
where
C = Phase correction offset.

But we don't need it for this problem, so the answer is:
f(x) = 5.05*sin(x*pi/12) + 5.15

Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>

7 0

3 years ago

Which type of light wave has the highest energy? which has the least amount of energy

devlian [24]

Answer:

Red has the lowest energy and violet the highest. Beyond red and violet are many other kinds of light our human eyes can't see, much like there are sounds our ears can't hear. On one end of the electromagnetic spectrum are radio waves, which have wavelengths billions of times longer than those of visible light.

Explanation:

6 0

3 years ago

HELP PLEASE!!

aliya0001 [1]

Answer:

m=41 kg

v=0,02 ms

R=2,1 m

F-?

a=v²/R

a=(0,02) ²/2,1=0,0002

F=m*a

F=41*0,0002=0,0082 H

F=0,0082 H

3 0

3 years ago

Prior Knowledge Questions (Do these BEFORE using the Gizmo.)A buoy is anchored to the ocean floor. A large wave approaches the b

dimaraw [331]

Answer:

bounce up and down

Explanation:

Buoys are used for two main reasons, one is to let the people on land know of a big incoming wave, while the second reason is to generate electricity. When a big wave is approaching the buoy starts to bounce up and down with the strength of the smalled previous waves and then bounce very strongly up as the bigger wave passes by. This movement is combined with pistons within the buoy in order to conduct electricity.

8 0

3 years ago