A baseball player throws a ball into the stands at 15.0 m/s and at an angle 45.0° above the horizontal. On its way down, the ball is caught by a spectator 4.10 m above the point where the ball was thrown. How much time did it take for the ball to reach the fan in the stands?

Answers

Answer 1
Answer:

Answer:

Time = 1.61 seconds

Explanation:

Using the equation displacement of a trajectory motion in the y plane

Y = u t sin ů - ½gt²....equation 1 where

Y= vertical displacement =4.1

U = initial velocity = 15m/s

g = acc. Due to gravity = 10m/s

Ů = angle of trajectory = 45

t = time to reach fan on its way down

Sub into equ 1

4.1 = 15t sinů - ½ * 10t²

4.1 = 10.61t - 5t²

Solve using quadratic formula

t =[-B±( -B² -4AC)^½]/2A....equation 2

Where A = 5, B=10.61, C =4.1

Substitute A,B,C into equ2

t = (10.61±5.53)/10

t = 0.508seconds or 1.61seconds

Since it is on its way down t= 1.61 seconds


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A bowling ball of mass 3 kg is dropped from the top of a tall building. It safely lands on the ground 3.5 seconds later. Neglecting air friction, what is the height of the building in meters? (Give the answer without a unit and round it to the nearest whole number)

Answers

The height of the building is 60 m.

calculation of building height:

The velocity of the ball should be provided by

v = u + gt

here,

u is the initial velocity of the ball = 0

v = 0 + 9.8 x 3.5

v = 34.3 m/s

Now

When the ball hits the ground, energy is conserved;

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (0.5 v²) / g

h = (0.5 x 34.3²) / (9.8)

h = 60.025 m

h = 60 m

Learn more about friction here: brainly.com/question/14455351

Answer:

The height of the building is 60 m.

Explanation:

Given;

mass of the mass of the ball, m = 3 kg

time of motion, t = 3.5 s

The velocity of the ball is given by;

v = u + gt

where;

u is the initial velocity of the ball = 0

v = 0 + 9.8 x 3.5

v = 34.3 m/s

When the ball hits the ground, energy is conserved;

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (0.5 v²) / g

h = (0.5 x 34.3²) / (9.8)

h = 60.025 m

h = 60 m

Therefore, the height of the building is 60 m.

*Which of the following cannot be an example of projectile motion
A. A football flying through the air
B. An apple falling from a tree
C. A pencil rolling on the ground
D.A rocket dropping from its maximum height

Answers

A. football flying through the air
A cause it flying through the also a projectile is a object flying in the air like a arrow for example/also can I get Brainly

Imagine that you drop an object of 10 kg, how much will be the acceleration andhow much force causes the acceleration?

Answers

If you do this on Earth, then the acceleration of the falling object is 9.8 m/s^2 ... NO MATTER what it's mass is.

If its mass is 10 kg, then the force pulling it down is 98.1 Newtons. Most people call that the object's "weight".

Air contains 78.08% nitrogen, 20.095% oxygen, and 0.93% argon. calculate the partial pressure of oxygen if the total pressure of the air sample is 1.7 atm.a.

Answers

partial pressure in a mixture of two or more gases will be given by formula

P_(partial) = mole fraction of gas * total pressure

now here mole fraction is same as percentage of gas in the mixture

Now mole fraction of oxygen is 0.20095 (20.095%)

now here pressure of oxygen in the mixture is given as

P_(o_2) = 0.20095 * P_(total)

P_(o_2) = 0.20095 * 1.7

P_(o_2) = 0.342 atm

so pressure due to oxygen in the mixture will be 0.342 atm

Answer:

20.095

Explanation:

A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The painter is standing 2 m from the first rope and 8 m from the second rope. What is the tension in the first rope (in Newtons)

Answers

Explanation:

For equilibrium, \sum M = 0.

So,   8 m * mg - (10 m) T_(1) = 0

             T_(1) = (8 * mg)/(10)

                        = (8 * 90 * 9.8)/(10)

                        = 705.6 N

Also, for equilibrium \sum F_(y) = 0

              T_(1) + T_(2) - mg = 0

or,         T_(2) = mg - T_(1)

                        = 90 * 9.8 - 705.6

                        = 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.

A 4 kg bowling ball accelerates 1 m/s2. what is the net force on the ball?

Answers

Answer:4N

Explanation:

mass=4kg

Acceleration=1m/s^2

Force=mass x acceleration

Force=4 x 1

Force=4N