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valkas [14]
3 years ago
14

If Galileo drops a cannon ball from the 60 meter high) Leaning Tower of Pisa, how fast will it be moving when it hits the ground

?
Physics
1 answer:
viva [34]3 years ago
5 0

Answer:

When the ball hits the ground, the velocity will be -34 m/s.

Explanation:

The height and velocity of the ball at any time can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

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

v = velocity at time "t".

If we place the origin of the frame of reference on the ground, when the ball hits the ground its height will be 0. Then using the equation of height, we can calculate the time it takes the ball to reach the ground:

y = y0 + v0 · t + 1/2 · g · t²

0 = 60 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

0 = 60 m - 4.9 m/s² · t²

-60 m / -4.9 m/s² = t²

t = 3.5 s

Now, with this time, we can calculate the velocity of the ball when it reaches the ground:

v = v0 + g · t

v = 0 m/s - 9.8 m/s² · 3.5 s

v = -34 m/s

When the ball hits the ground, the velocity will be -34 m/s.

You might be interested in
A 1000.0 kg truck accelerates from 20.0 m/s to 25.0 m/s over a distance of 300.0 m. What is the average net force on the truck?
choli [55]

Answer:

The average net force on the truck is 375 Newtons.

Explanation:

Using Newton's 3rd equation of motion, we have :

v^{2} - u^{2} = 2×a×s

where, v = final velocity = 25 m/s

u = initial velocity = 20 m/s

a = acceleration

s = distance traveled = 300 m

Using these values in the above equation, we get acceleration = 0.375 m/s^{2}

Using Newton's second law, we have:

F=m×a

where m = mass = 1000 kg

a= acceleration = 0.375 m/s^{2}

Putting values we have F=375 N

3 0
3 years ago
Suppose a woman raises a 65 N object in 2m in 4 seconds.
Novosadov [1.4K]

Answer:

\huge\boxed{\sf P.E = 130\ Joules}

\huge\boxed{\sf P = 32.5\ Watts}

Explanation:

<u>Given Data:</u>

Weight = W = 65 N

Height = h = 2 m

Time = t = 4 secs

<u>Required:</u>

Power = P = ?

Work Done in the form of Potential Energy = P.E. = ?

<u>Formula:</u>

P.E. = Wh

P = P.E. / t

<u>Solution:</u>

P.E. = (65)(2)

P.E = 130 Joules

P = P.E. / t

P = 130 / 4

P = 32.5 Watts

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807 </h3>
8 0
3 years ago
A 50 kg pitcher throws a baseball with a mass of 0. 15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is
dsp73

The velocity of the pitcher at the given mass is 0.1 m/s.

The given parameters:

  • <em>Mass of the pitcher, m₁ = 50 kg</em>
  • <em>Mass of the baseball, m₂ = 0.15 kg</em>
  • <em>Velocity of the ball, u₂ = 35 m/s</em>

<em />

Let the velocity of the pitcher = u₁

Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;

m₁u₁ = m₂u₂

u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.15 \times 35}{50} \\\\u_1 = 0.105 \ m/s\\\\u_1 \approx 0.1 \ m/s

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/13589460

4 0
2 years ago
I’m designing a kitchen for a person in a wheel chair, which accommodation is least important? A. Provide turn around space for
Valentin [98]
The least important accommodation would be C. 
7 0
3 years ago
Read 2 more answers
A fully loaded cart with a mass of 2200 kg starts from the top of a 12-meter hill on a roller coaster.
Salsk061 [2.6K]

Answer:

A. potential energy is 258720 Joule

Explanation:

A.Gravitational potential energy is: PE = m × g × h

velocity =  15.33 m/s when the car reaches the bottom of the hill.

where, m = mass

            g = acceleration due to gravity

            h = height from the bottom of hill.

The potential energy is : m×g×h

                                     =(2200×9.8×12)

                                     =258720 Joule

B. at the bottom of the hill, the potential energy is converted into kinetic energy so PE at top = KE at bottom

                    kinetic energy= \frac{1}{2}(m*v^{2})

where v = velocity

          m= mass

therefore,               v=\sqrt\frac{2*K.E}{m} {}

                         or,  v=\sqrt{\frac{2*258720}{2200} }

                         or,   v=15.33 m/s

7 0
3 years ago
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