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borishaifa [10]
3 years ago
13

Four seconds after being launched, what is the height of a ball that starts from a height of 12 m with an initial upward velocit

y of 24 m/s?
Physics
1 answer:
MrMuchimi3 years ago
8 0

Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

Assuming that the upward direction is positive.

As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.

Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.

Given that u=24 m/s, t=4 seconds, g=-9.81 m/s^2.

So, the final velocity is

v= 24 + (-9.81)\times 4 \\\\\Rightarrow v= 24-9.81\times 4

\Rightarrow v=-15.24 m/s

Here, the negative sign means the final velocity is in the downward direction.

Hence, the velocity after 4 seconds is 15.24 m/s in the downward direction.

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You are riding a bicycle at 20 m/s at 10:00 am. At 10:01 am, you notice that you are traveling at 40 m/s. You are headed W. What
Juliette [100K]

Answer:

20m/s^2

Explanation:

Acceleration=Change in velocity/time taken for change

40-20/1

20m/s^2

7 0
3 years ago
4) A cannon shoots a cannonball of mass 5kg vertically upward from the mouth of a cannon with muzzle velocity of 7 m/s. At a hei
Ad libitum [116K]

Answer:

h =220 m

Explanation:

Given that

u = 7 m/s

Even mass will attach but this will not produce any effect on the maximum height of the ball.Because in energy conservation the effect of mass does not present.

So the final speed of the ball will be zero at the maximum height.

v² = u² - 2 g (25 + h)

0 = 7² - 2 x 10 (25 +h)

49 = 20 ( 25 +h)

49 = 500 +20 h

Here h comes out negative that is why we are taking the 70 m/s in place of 7 m/s.

0 = 70² - 2 x 10 (25 +h)         ( take g =10 m/s²)

4900 = 20 ( 25 +h)

4900 = 500 +20 h

4900- 500 = 20 h

4400 = 20 h

440 = 2 h

h =220 m

5 0
3 years ago
A bullet is shot vertically upward with an initial velocity of 128 ft/s. The bullet's height after t seconds is y(t) = 128t - 16
saul85 [17]

The height of the bullet when the velocity is zero is 256 ft.

<h3>Height of the bullet when the velocity is zero </h3>

The height of the bullet when the velocity is zero is determined by taking derivative of the function as shown below;

v = \frac{dy}{dt} = 128 -32t \\\\when \ v \ is \ zero\\\\v = 0\\\\128 - 32t = 0\\\\32t = 128\\\\t = \frac{128}{32} \\\\t = 4 \  s

The height of the bullet at this time is calculated as follows;

y(4) = 128(4) - 16(4)^2\\\\y(4) = 256 \ ft

Learn more about height of projectiles here: brainly.com/question/10008919

6 0
2 years ago
When the net force of opposite forces is zero , the forces are
lorasvet [3.4K]
The answer is balanced
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3 years ago
A spelunker is surveying a cave. She follows a passage 140 m straight west, then 290 m in a direction 45? east of south, and the
zimovet [89]

Answer:

Magnitude of fourth displacement is approximately 95 metres,

Direction of fourth displacement is straight west.

3 0
3 years ago
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