PHYSICS COLLEGE

A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

Answers

Answer 1

Answer:

Answer:

22.1 m

Explanation:

$v_(o)$ = initial speed of ball = 14.3 m/s

$\theta$ = Angle of launch = 27°

Consider the motion of the ball along the vertical direction.

$v_(oy)$ = initial speed of ball = $v_(o) Sin\theta = 14.3 Sin27 = 6.5 ms^(-1)$

$a_(y)$ = acceleration due to gravity = - 9.8 ms⁻²

$t$ = time of travel

$y$ = vertical displacement = - 3.50 m

Using the kinematics equation that suits the above list of data, we have

$y = v_(oy) t + (0.5) a_(y) t^(2) \n- 3.50 = (6.5) t + (0.5) (- 9.8) t^(2)\n- 3.50 = (6.5) t - 4.9 t^(2)\nt = 1.74 s$

Consider the motion of the ball along the horizontal direction.

$v_(ox)$ = initial speed of ball = $v_(o) Cos\theta = 14.3 Cos27 = 12.7 ms^(-1)$

$X$ = Horizontal distance traveled

$t$ = time taken = 1.74 s

Since there is no acceleration along the horizontal direction, we have

$X = v_(ox) t\nX = (12.7)(1.74)\nX = 22.1 m$

Related Questions

An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless spring, compressing it a maximumdistance of 0.750m. (See below.) The spring constant is 500N/m, the height of the incline is 2.0 m, and the horizontalsurface is frictionless. (a) What is the speed of the object atthe bottom of the incline? (b) What is the work of frictionon the object while it is on the incline? (c) The springrecoils and sends the object back toward the incline. Whatis the speed of the object when it reaches the base of theincline? (d) What vertical distance does it move back up theincline?
Express the following speeds as a function of the speed of light, c: (a) an automobile speed (93 km/h) (b) the speed of sound (329 m/s) (c) the escape velocity of a rocket from the Earth's surface (12.1 km/s) (d) the orbital speed of the Earth about the Sun (Sun-Earth distance 1.5×108 km).
A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in the +z direction exists throughout the plane of the loop. The current has the magnitude of I and it is deirected counterclockwise when observing from positive z axis.What is the magnetic force exerted by the external field on the loop?Express your answer in terms of some or all of the variables I, R, and B
You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75 m wide and 1.5 m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower? These frequencies will be especially amplified. Assume a sound speed of 343 m/s.
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)

A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at x = 2.00 m is 1 000 V, what is the change in potential energy of a particle with a charge of + 2.80 x 10-3 C as it moves from x = 1.9 m to x = 2.1 m?

Answers

Answer:

$\Delta U = 0.2072 J$

Explanation:

Potential difference between two points in constant electric field is given by the formula

$\Delta V = E.\Delta x$

here we know that

$E = 370 N/C$

also we know that

$\Delta x = 2.1 - 1.9 = 0.2 m$

now we have

$\Delta V = 370 (0.2) = 74 V$

now change in potential energy is given as

$\Delta U = Q\Delta V$

$\Delta U = (2.80 * 10^(-3))(74)$

$\Delta U = 0.2072 J$

The cycle is a process that returns to its beginning, but it does not repeatitself.
True
False

Answers

False. It does repeat itself

A baseball is hit with a speed of 47.24 m/s from a height of 0.42 meters. If the ball is in the air 5.73 seconds and lands 130 meters from the batters feet, (a) at what angle did the ball leave the bat? (b) with what velocity will the baseball hit the ground?

Answers

Answer:

a)the ball will leave the bat at an angle of 61.3° .

b) the velocity at which it will hit the ground will be v = 27.1 m/s

Explanation:

Given,

v = 47.24 m

h = 0.42 m

t = 5.73 s

R = 130 m

a)We know that

R = v cosθ × t

cosθ = $(R)/(v t ) = (130)/(47.24* 5.73 ) =0.4803$

θ = 61.3°

the ball will leave the bat at an angle of 61.3° .

b)Vx = v cos(θ) = 47.24 x cos 61.3 = 22.7 m/s

v = u + at

Vy = 47.24 x sin 61.3 - 9.81 x 5.73

= -14.8 m/s

v = $√(v_x^2 + v_y^2))$

v = $√(22.7^2 + -14.8^2)$

v = 27.1 m/s

the velocity at which it will hit the ground will be v = 27.1 m/s

Captain Kiddo is trapped inside a submarine. In order to escape, she first needs to open a heavy revolving door. a. Would be easier for her to apply all her force close to the axle of the door or as far away as possible from it?
b. To finally get the hatch open she needs to grab hold of the wheel with her hands on opposite sides and rotate the wheel by exerting a force toward the top of the door on one side while exerting the same force towards the bottom of the door on the opposite side. Is there a net force applied on the wheel? Is there a net torque applied? Explain your answers

Answers

Answer:

a) It will be easier to apply all her force as far as possible from the axle

b) There is a net torque applied

Explanation:

a) Applying a larger force far away from the axle of the door produces a larger force on the torque than pushing it near the axle. This is because Increasing the lever arm between the axle and the point at which you push the door increases the torque on the door

b)The door is turning faster and faster because you are exerting a torque on it and its undergoing angular acceleration. There is a net torque which is the addition of the torque applied on the opposite sides of the door

Charge q1 is placed a distance r0 from charge q2 . What happens to the magnitude of the force on q1 due to q2 if the distance between them is reduced to r0/4 ?What is the electrostatic force between and electron and a proton separated by 0.1 mm?

Answers

Answer:

The electrostatic force between and electron and a proton is $F=2.30* 10^(-20)\ N$

Explanation:

It is given that, charge $q_1$ is placed at a distance $r_o$ from charge $q_2$ . The force acting between charges is given by :

$F=(kq_1q_2)/(r_o^2)$

We need to find the force if the distance between them is reduced to $r_o/4$ . It is given by :

$F'=(kq_1q_2)/((r_o/4)^2)$

$F'=16* (kq_1q_2)/(r_o^2)$

$F'=16* F$

So, if the the distance between them is reduced to $r_o/4$ , the new force becomes 16 times of the previous force.

The electrostatic force between and electron and a proton separated by 0.1 mm or $10^(-4)\ m$ is :

$F=(kq_1q_2)/(r_o^2)$

$F=(9* 10^9* (1.6* 10^(-19))^2)/((10^(-4))^2)$

$F=2.30* 10^(-20)\ N$

So, the electrostatic force between and electron and a proton is $F=2.30* 10^(-20)\ N$ . Hence, this is the required solution.

A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, `A D 0:28 m, and `B D 0:36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.