All categories

3 years ago

You throw a ball straight up. You are extra strong feeling today. It takes 11 seconds for the ball to come back down.

Physics

1 answer:

ollegr [7]3 years ago

5 0

Answer:

A. 148.23 m

B. 2.75 m/s

Explanation:

The following data were obtained from the question:

Time of flight (T) = 11 s

Maximum height (h) =?

Initial velocity (u) =?

Next, we shall determine the time taken for the ball to get to the maximum height. This can be obtained as follow:

Time of flight (T) = 11 s

Time (t) to reach the maximum height =.?

T = 2t

11 = 2t

Divide both side by 2

t = 11/2

t = 5.5 s

NOTE: Time to reach the maximum height is the same as the time taken for the ball to fall back to the plane of projection.

A. Determination of the maximum height to which the ball was thrown.

Time (t) to reach maximum height = 5.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =?

h = ½gt²

h = ½ × 9.8 × 5.5²

h = 4.9 × 30.25

h = 148.23 m

B. Determination of the initial velocity.

Maximum height (h) reached = 148.23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

u² = h/2g

u² = 148.23 / (2 × 9.8)

u² = 148.23 / 19.6

Take the square root of both side

u = √(148.23 / 19.6)

u = 2.75 m/s

You might be interested in

A space station has a mass M and orbits Earth in a circular orbit at a height above Earth’s surface. An astronaut in the space s

Svetllana [295]

Answer:

The correct option is;

D) The force exerted on the astronaut by Earth is equal to the force exerted on Earth by the astronaut

Explanation:

According to Newton's third law of motion, in nature, for every action, there is an equal and opposite reaction, such that if a first object exerts a certain amount of force on a second object, the second object will exert a force of equal magnitude and opposite direction to that exerted by the first object

Therefore, the gravitational force exerted by Earth on the astronaut, is equal to the force exerted by the astronaut on Earth.

8 0

3 years ago

The point on the graph that lies on the y-axis (vertical axis) is called the y-intercept. What does the y-intercept tell you abo

jekas [21]

Answer:

The starting position of the runner.

Explanation:

When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.

The runner starts at twenty, and ends at thirty.

Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.

7 0

3 years ago

Lewis structure for BeCI2<br>

ludmilkaskok [199]

Answer:

you can see it in the picture

5 0

2 years ago

A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790

Nezavi [6.7K]

Answer:

$w_2=0.467rad/s$

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

$m_t=4*70kg=280kg$

The radius $r=4.2/2=2.1m$

Angular velocity $w_1=0.79rad/s$

The moment of inertia total is $I_t=1790 kg/m^2$

Momento if inertia

$I_1=m_t*r^2$

$I_1=280kg*(2.1m)^2=1234.8kg*m^2$

Angular momentum

$I_1*w_1=I_t*w_2$

Solve to w2

$w_2=\frac{I_1*w_1}{I_t}$

$w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}$

$w_2=0.467rad/s$

8 0

3 years ago

Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist

andreyandreev [35.5K]

Answer:

a) It is moving at $43.15\frac{m}{s^{2}}$ when reaches the ground.

b) It is moving at $101.44\frac{m}{s^{2}}$ when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

$W=K_f-K_i$ (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with m the mass and v the velocity.

Using (2) on (1):

$W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

$W=Fd=wd=mgd$ (4)

Using (4) on (3):

$mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

$v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}$

$v_f=43.15\frac{m}{s^{2}}$

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

$-mgd=-\frac{mv_i^2}{2}$

Solving for initial velocity (when the boulder left the volcano):

$v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}$

$v_i=101.44 \frac{m}{s^{2}}$

3 0

3 years ago