Ask question

Ask question

All categories

lorasvet [3.4K]

3 years ago

12

A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come

back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground. (a) How fast was the ball originally moving when it was kicked. (b) How much longer would it take the ball to reach the ground?

1 answer:

Yuliya22 [10]3 years ago

5 0

Answer:

(a) vo = 24.98m/s

(b) t = 5.09 s

Explanation:

(a) In order to calculate the the initial speed of the ball, you use the following formula:

$y=y_o+v_ot-\frac{1}{2}gt^2$ (1)

y: vertical position of the ball = 2.44m

yo: initial vertical position = 0m

vo: initial speed of the ball = ?

g: gravitational acceleration = 9.8m/s²

t: time on which the ball is at 2.44m above the ground = 5.00s

You solve the equation (1) for vo and replace the values of the other parameters:

$v_o=\frac{y-y_o+1/2gt^2}{t}$

$v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}$

The initial speed of the ball is 24.98m/s

(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:

$0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s$

The time that the ball takes to arrive to the ground is 5.09s

You might be interested in

A portable basketball set has a base and a post arrangement. The post arrangement consists of a post, backboard, hoop and net. T

Ray Of Light [21]

The rotational equilibrium condition allows finding the response to the minimum force of the wind and what happens when changing the water for sand, in the system

a) The minimum force of the wind that turns the system is Fw = 17.64 N

b) The system resists much greater forces because the base has more mass

Newton's Second Law can be applied to rotational motion in this case when the angular acceleration is zero we have the special case of rotational equilibrium

Σ τ = 0

Where τ is the torque

The reference system is a coordinate system with respect to which the torques are measured, in this case we will fix the system at the turning point, the junction of the base and the pole, we will assume that the counterclockwise rotations are positive.

For the torque the distance used is the perpendicular distance from the direction of the force to the axis of rotation, let's find this distance for each force

Wind force

cos 15 = $\frac{y_w}{2.35}$

$y_w$ = 2.35 cos 15

Post Weight

sin 15 = $\frac{x_p}{2.00}$

xp = 2.0 sin 15

Base weight

cos (90-15) = $\frac{x_b}{0.25}$

xB = 0.25 cos 75

Let's substitute in the rotational equilibrium equation

$F_w \ y_w + W_p \ x_p - W_b \ x_b = 0$

a) To calculate the minimum wind force we substitute the given values

They indicate the weight of the post is $W_p$ = 26.0 N and the weight of the base with water is $W_b$ = 810 N

$F_w = \frac{W_b \ x_b - W_p \ x_p }{y_w}$

$F_w = \frac{W_b \ 0.25 cos75 \ - W_p \ 2 sin 15}{2.35 cos 15}$

Let's calculate

$F_w = \frac{810 \ 0.25 \ cos75 \ - 26.0 \ 2 \ sin 15}{2.35 cos15}\\F_w = \frac{52.41 - 10.30}{2.3699}$

$F_w$ = 17.64 N

b) The water is exchanged for sand.

In this case, as the density of the sand is greater than that of the water, the base will have more weight, so it will resist stronger winds before turning over.

Using the rotational equilibrium condition we can find the response to the minimum force of the wind and what happens when changing the water for sand,

a) the minimum force of the wind that turns the system is Fw = 17.64 N

b) the system resists much greater forces because the base has more mass

Learn more here: brainly.com/question/7031958

3 0

3 years ago

The energy source used to produce most of the electrical energy in the united states is

quester [9]

Coal power, using fossil fuels.

5 0

3 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

6 0

3 years ago

A space probe is orbiting a planet on a circular orbit of radius R and a speed v. The acceleration of the probe is a. Suppose ro

Mamont248 [21]

Answer:

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

$a_{new} = 8a.$

Explanation:

given data:

radius of orbit = R

Speed pf planet = v

new radius = 0.5R

new speed = 2v

we know that acce;ration is given as

$a = \frac{v^{2}}{R},$

$a_{new} =\frac{(2v)^{2}}{0.5R},$

$= \frac{4v^{2}}{0.5R}$

$= \frac{8 v^{2}}{R}$

$a_{new} = 8a.$

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

7 0

4 years ago

find the distance traveled in km when the average speed is 60 km per hour and the time taken is 3 and 1/2 hours

Anna35 [415]

Answer:

210 km

Explanation:

60 x 3 1/2 = 210

3 0

3 years ago