Madelin fires a bullet horizontally. The rifle is 1.4 meters above the ground. The bullet travels 168 meters horizont before it hits the ground. What speed did Madelin's bullet have when it exited the rifle?


Answer 1

The position vector of the bullet has components


y=1.4\,\mathrm m-\frac g2t^2

The bullet hits the ground when y=0, which corresponds to time t:

1.4\,\mathrm m-\frac g2t^2=0\implies t=0.53\,\mathrm s

The bullet travels 168 m horizontally, which would require a muzzle velocity v_0 such that

168\,\mathrm m=v_0(0.53\,\mathrm s)

\implies v_0\approx320\,(\mathrm m)/(\mathrm s)

Answer 2

Final answer:

In the given physics problem, the bullet travels horizontally 168 meters before hitting the ground from a height of 1.4 meters. By calculating the time it takes for the bullet to fall to the ground due to gravity and then applying that time to the horizontal distance traveled, we find that the speed of the bullet when it exited the rifle was approximately 313.43 m/s.


The scenario defined is a classic Physics problem where an object is fired horizontally and falls to the ground due to gravity. We can calculate the horizontal speed of the bullet using the equations of motion associated with the vertical, free-fall motion of the bullet.

Gravity causes the bullet to fall to the ground. As we know that the height from the ground is 1.4 meters, we can calculate the time taken for the bullet to hit the ground using the equation: time = sqrt(2 * height / g), where g is the gravitational constant (approx. 9.8 m/s^2).

Substituting the given value, we get time = sqrt(2 * 1.4 / 9.8), which is around 0.536 seconds. The bullet travels 168 meters in this time horizontally, therefore its horizontal speed will be distance / time, which is 168 meters / 0.536 seconds = 313.43 m/s. So, Madelin's bullet had a speed of around 313.43 m/s when it exited the rifle.

Learn more about Projectile Motion here:


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In this example the motion is in a vertical circle. Passengers in a carnival ride travel in a circle with radius 5.0 m. The ride moves at a constant speed and makes one complete circle in a time T=4.0s. What is the acceleration of the passengers? If the ride increases in speed so that T=3.0s, what is arad? (This question can be answered by using proportional reasoning, without much arithmetic.)



a. 12.3m/s^(2)

b. 21.93m/s^(2)


From the data given, the radius is 5.0m, and the time taken to complete one circle is 4.0secs

Since the motion is in a circular part, we can conclude that the total distance covered in this time is given as circumference of the circle.

which is expressed as

Distance=2\pi R

To determine the speed, we use the equation

speed=(distnce)/(time)\n Speed=(2\pi R)/(time)\n speed=(2\pi*5 )/(4)\n Speed=7.85m/s

The acceleration as required is expressed as

a=(v^(2))/(r)\n a=(7.85^(2))/(5)\n a=12.3m/s^(2)

if the speed increase and it takes 3secs to complete one circle, the speed is

speed=(distnce)/(time)\n Speed=(2\pi R)/(time)\n speed=(2\pi*5 )/(3)\n Speed=10.47m/s

and the acceleration becomes

a=(v^(2))/(r)\n a=(10.47^(2))/(5)\n a=21.93m/s^(2)

The acceleration of the passengers in the vertical circle carnival ride is 19.6 m/s^2. When the time taken to complete one circle is 3.0 s, the new acceleration is 26.13 m/s^2.

The acceleration of the passengers can be determined using the centripetal acceleration formula, which is given by a = v^2 / r.

In this case, the velocity v can be found by dividing the circumference of the circle (2πr) by the time taken to complete one circle (T). The radius r is given as 5.0 m. Plugging in the values, we have:

a = (v^2) / r = ((2πr / T)^2) / r = (4π^2r) / T^2 = (4π^2 * 5.0) / 16.0 = 19.6 m/s^2

To find the new acceleration when the time taken to complete one circle is 3.0 s, we can use the proportional reasoning to determine the relationship between the two accelerations. Since the time is inversely proportional to the acceleration, when T is 3.0 s, the new acceleration arad can be found using the equation:

arad / 19.6 = 4.0 / 3.0

Simplifying the equation, arad = (19.6 * 4.0) / 3.0 = 26.13 m/s^2

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At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.60 m/s^2 . At the same instant a truck, traveling with a constant speed of 23.5 m/s , overtakes and passes the car. a. How far beyond its starting point does the car overtake the truck?b. How fast is the car traveling when it overtakes the truck?



306.8264448 m

47.0016 m/s


t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled by car

s_c=ut+(1)/(2)at^2\n\Rightarrow s_c=(1)/(2)at^2

Distance traveled by truck


In order to overtake both distances should be equal

(1)/(2)at^2=ut\n\Rightarrow (1)/(2)at=u\n\Rightarrow t=(2u)/(a)\n\Rightarrow t=(2* 23.5)/(3.6)\n\Rightarrow t=13.056\ s

s_c=(1)/(2)at^2\n\Rightarrow s_c=(1)/(2)3.6* 13.056^2\n\Rightarrow s_c=306.8264448\ m

The distance the car has to travel is 306.8264448 m

v^2-u^2=2as\n\Rightarrow v=√(2as+u^2)\n\Rightarrow v=√(2* 3.6* 306.8264448+0^2)\n\Rightarrow v=47.0016\ m/s

The speed of the car when it overtakes the truck is 47.0016 m/s

A fly lands on one wall of a room. The lower-left corner of the wall is selected as the origin of a two-dimensional Car- tesian coordinate system. If the fly is located at the point having coordinates (2.00, 1.00) m, (a) how far is it from the origin? (b) What is its location in polar coordinates?



a) The fly is 2.24 m from the origin.

b) In polar coordinates, the position of the fly is (2.24 m, 26.7°).


Hi there!

The position vector of the fly is r = (2.00, 1.00)m. The distance from that point to the origin is the magnitude of the vector "r" (see figure).

a) Notice in the attached figure that the distance from the origin to the point where the fly is located is the hypotenuse of the triangle formed by r, the x-component of r (2.00 m) and the y-component ( 1.00 m). Then:

r² = (2.00 m)² + (1.00 m)²

r² = 5.00 m²

r = 2.24 m

The fly is 2.24 m from the origin.

b) To find the angle θ (see figure) we can use trigonometry:

cos θ = adjacent / hypotenuse

cos θ = 2.00 m / √5 m

θ = 26.7°

The same will be obtained if we use sin θ:

sin θ = opposite / hypotenuse

sin θ = 1.00 m / √5 m

θ = 26.7°

In polar coordinates, the position of the fly is (2.24 m, 26.7°).

A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?





According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis,  and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

\int\ {E} \, dA = E*A = E*\pi  *R^(2)

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

Flux = E*π*R²

Of the four most important pathways by which stress affects health, the first one to occuris usually related to physiology


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