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insens350 [35]
4 years ago
13

A batter hits a pitched ball when the center of the ball is 1.22 m above the ground. The ball leaves the bat at an angle of 45°

with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107 m.
(a) Does the ball clear a 7.32-m-high fence that is 97.5 m horizontally from the launch point?
(b) At the fence, what is the distance between the fence top and the ball center?

Physics
1 answer:
Sveta_85 [38]4 years ago
7 0

Answer:

2.56m

Explanation:

This is problem 4.47 from your textbook.

Find: Whether a batted baseball clears a fence, and by what amount it does or does not.

Given: The baseball’s initial launch height and angle, the range the baseball would have without the fence, the distance to the fence and its height.

Let the y axis run vertically and the x axis horizontally. Let the range the baseball would have without the fence be R=107 m, with the distance to the fence d=97.5m and its height hfence=7.32 m. The baseball is batted at an angle θ=45° at speed vi a height of hbat=1.22m above the ground.

Let the origin be at the position the ball leaves the bat. The height of the fence relative to the height of the bat is then

δh = hfence − hbat

What we really need to determine is the ball’s y coordinate at x = d. If y > δh, the ball clears the fence. We can use the range the baseball would have without the fence and the launch angle to find the ball’s speed, which will allow a complete calculation of the trajectory.

Relevant equations: We need only the equations for the range and trajectory of a projectile over level ground:

R = (vi*sin2θ)/g

For convenience sake and easy reading, I extracted the solution of your textbook for the remaining parts of the solution.

Hence, it is seen that the ball does clear the fence, by approximately 2.56 m

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7 0
3 years ago
A 4 kg and 6 kg bowling ball are dropped from the same height at the same time. The two balls strike the ground at the same time
Ray Of Light [21]
The force of gravity on objects is proportional to the mass of each object.

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5 0
3 years ago
Read 2 more answers
A 1100 kgkg safe is 2.4 mm above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compres
Drupady [299]

Answer:

191.36 N/m

Explanation:

From the question,

The Potential Energy of the safe = Energy of the spring when it was compressed.

mgh = 1/2ke²............... Equation 1

Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression

Making k the subject of the equation,

k =2mgh/e²................ Equation 2

Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m

Constant: g = 9.8 m/s²

Substitute into equation 2

k = 2(1100)(9.8)(0.0024)/0.52²

k = 51.744/0.2704

k = 191.36 N/m

Hence the spring constant of the heavy-duty spring = 191.36 N/m

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3 years ago
A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a spe
Aleksandr-060686 [28]

T<u>he direction of motion</u> of the person relative to the water is <u>16.7° north of east.</u>

Why?

We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)

We can calculate the angle that give the direction using the following formula:

Tan(\alpha)=\frac{NorthSpeed}{EastSpeed}\\\\Tan(\alpha)^{-1}=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}\\\\\alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}

Now, substituting the given information we have:

alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}

\alpha =Tan(\frac{3.6\frac{m}{s} }{12\frac{m}{s} })^{-1}\\\\\alpha =Tan(0.3)^{-1}=16.69\°(North-East)=16.7\°(North-East)

Hence, we have that <u>the direction of motion</u> of the person relative to the water is 16.7° north of east.

Have a nice day!

7 0
3 years ago
A robot is exploring​ charon, the dwarf planet​ pluto's largest moon. Gravity on charon is 0.278 meters per second ​[m divided b
DochEvi [55]

In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up

First we will find the mass of the stone

As it is given that stone is spherical in shape so first we will find its volume

V = \frac{4}{3}\pi r^3

V = \frac{4}{3}\pi *(\frac{0.06}{2})^3

V = 1.13 * 10^{-4} m^3

Now it is given that it's specific gravity is 10.8

So density of rock is

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mass of the stone will be

m = \rho V

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m = 1.22 kg

now change in potential energy is given as

\Delta U = mgH

here

g = gravity on planet = 0.278 m/s^2

H = height lifted upwards = 15 cm

\Delta U = 1.22* 0.278 * 0.15

\Delta U = 0.051 J

Now energy supplied by internal circuit of robot is given by

E = Vit

V = voltage supplied = 10 V

i = current = 1.83 mA

t = time = 12 s

E = 10* 1.83 * 10^{-3} * 12

E = 0.22 J

Now efficiency is defined as the ratio of output work with given amount of energy used

\eta = \frac{\Delta U}{E}*100

\eta = \frac{0.051}{0.22} = 0.23

so efficiency will be 23 %

5 0
3 years ago
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