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3 years ago

A ball is thrown horizontally from a 90 m cliff and strikes the ground 70 m from the base, what is the initial velocity?

Physics

1 answer:

mash [69]3 years ago

7 0

Answer:

The initial velocity is 16 m/s

Explanation:

The position of the ball at a time t is given by the following vector:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector at time t

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

When the ball reaches the ground its position vector will be "r final" (see attached figure). The x-component of this vector is 70 m and the y-component is -90 m. Notice that the origin of the frame of reference is located at the throwing point (y0 and x0 = 0). Then, we can use the equations for the x and y-component of "r" to obtain the time of flight and the initial velocity:

Using the equation for the y-component of the vector position:

y = y0 + v0y · t + 1/2 · g · t² (y0 and v0y = 0)

-90 m = -1/2 · 9.8 m/s² · t²

-2 · -90 m/ 9.8 m/s² = t²

t = 4.3 s

Using the equation for the x-component of the vector position:

x = x0 + v0x · t

70 m = v0x · 4.3 s

v0x = 16 m/s

The initial velocity is 16 m/s

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Standing on the roof of a (42.0+A) m tall building, you throw a ball straight up with an initial speed of (14.5+B) m/s. If the b

RSB [31]

First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:

2as = v² - u²

Because the final velocity v is 0 in such cases

s = -u²/2a; because both u and a are downwards, the negative sign cancels

s = 14.5² / 2*9.81
s = 10.72 meters

Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m

We will use the formula
s = ut + 0.5at²

to find the time taken with the initial velocity u = 0.

55.72 = 0.5 * 9.81 * t²

t = 3.37 seconds

4 0

3 years ago

The moment of inertia for a 5500 kg solid disc is 12100 kg-m^2. Find the radius of the disc? (a) 2.111 m (b) 2.579 m (c) 1.679

Soloha48 [4]

Answer:

The radius of the disc is 2.098 m.

(e) is correct option.

Explanation:

Given that,

Moment of inertia I = 12100 kg-m²

Mass of disc m = 5500 kg

Moment of inertia :

The moment of inertia is equal to the product of the mass and square of the radius.

The moment of inertia of the disc is given by

$I=\dfrac{mr^2}{2}$

Where, m = mass of disc

r = radius of the disc

Put the value into the formula

$12100=\dfrac{5500\times r^2}{2}$

$r=\sqrt{\dfrac{12100\times2}{5500}}$

$r= 2.098\ m$

Hence, The radius of the disc is 2.098 m.

8 0

3 years ago

At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl

Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

$s=ut+0.5at^2 \\$

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

$h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9$

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

$h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\$

Solving both equations

$600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\$

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

$h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\$

Passing of B occurs at 4108.31 height.

6 0

3 years ago

At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v

Cerrena [4.2K]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.

$R= 6370*10^3 m$