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2 years ago

A ball is in free fall after being dropped. What willthe speed of the ball be after 2 seconds of free fall?

Physics

1 answer:

Mazyrski [523]2 years ago

8 0

So, the speed of the ball after 2 seconds after free fall is <u>20 m/s</u>.

<h3>Introduction</h3>

Hi ! I'm Deva from Brainly Indonesia. In this material, we can call this event "Free Fall Motion". There are two conditions for free fall motion, namely falling (from top to bottom) and free (without initial velocity). Because the question only asks for the final velocity of the ball, in fact, we may use the formula for the relationship between acceleration and change in velocity and time. In general, this relationship can be expressed in the following equation :

$\boxed{\sf{\bold{a = \frac{v_2 - v_1}{t}}}}$

With the following conditions :

a = acceleration (m/s²)
$\sf{v_2}$ = speed after some time (m/s)
$\sf{v_1}$ = initial speed (m/s)
t = interval of time (s)

<h3>Problem Solving</h3>

We know that :

a = acceleration = 9,8 m/s² >> because the acceleration of a falling object is following the acceleration of gravity (g).
$\sf{v_1}$ = initial speed = 0 m/s >> the keyword is free fall
t = interval of time = 2 s

What was asked :

$\sf{v_2}$ = speed after some time = ... m/s

Step by step :

$\sf{a = \frac{v_2 - v_1}{t}}$

$\sf{(a \times t) + v_1 = v_2}$

$\sf{(10 \times 2) + 0 = v_2}$

$\boxed{\sf{v_2 = 20 \: m/s}}$

So, the speed of the ball after 2 seconds after free fall is 20 m/s.

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3 years ago

Read 2 more answers

A rocket engine uses fuel and oxidizer in a reaction that produces gas particles having a velocity of 1380 ms The desired thrust

bearhunter [10]

Answer:

a. 141.3 kg/s b. 5.49 m/s² c. i. 104228.9 N ii. 8.53 m/s² d. i. 97305.2 N ii. 9.84 m/s²

Explanation:

a. What must be the fuel/oxidizer consumption rate (in kg s1)?

The thrust T = Rv where R = mass consumption rate and v = velocity of rocket. Since T = 195000 N and v = 1380 m/s,

R = T/v = 195000 N/1380 m/s = 141.3 kg/s

b. If the initial weight of the rocket is 125000 N, what is its initial acceleration?

We also know that thrust T - W = ma since the rocket has to move against gravity. where M = mass of rocket = W/g = 125000 N/9.8m/s² = 12755.1 kg, W = weight of rocket = 125000 N, a = acceleration of rocket and T = thrust = 195000 N.

So, T - W = Ma

195000 N - 125000 N = (12755.1 kg)a

70000 N = ma

a = 70000 N/12755.1 kg = 5.49 m/s²

c. What are the weight and acceleration of the rocket at t 15.0 s after ignition?

We know that the loss in mass ΔM = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 15 s,

ΔM = 141.3 kg/s × 15 = 2119.5 kg

The new mass is thus M = M - ΔM = 12755.1 kg - 2119.5 kg = 10635.6 kg

i.The weight after 15 seconds is thus W' = M'g = 10635.6 kg × 9.8m/s² = 104228.9 N

ii. Since T - W' = M'a. where M' is our new mass and a our new acceleration,

a = (T - W')/M'

= (195000 N - 104228.9 N)/10635.6 kg

= 90771.1 N/10635.6 kg

= 8.53 m/s²

d. What are the weight and acceleration of the rocket at 20.0 s after ignition?

We know that the loss in mass ΔM" = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 20 s,

ΔM" = 141.3 kg/s × 20 = 2826 kg

The new mass is thus M" = M - ΔM" = 12755.1 kg - 2826 kg = 9929.1 kg

i. The weight after 20 seconds is thus W" = M"g = 9929.1 kg × 9.8m/s² = 97305.2 N

ii. Since T - W" = M"a. where M" is our new mass and a our new acceleration,

a = (T - W")/M"

= (195000 N - 97305.2 N)/9929.1 kg

= 97694.8 N/9929.1 kg

= 9.84 m/s²

4 0

3 years ago

A 64 kg swimmer jumps, with a velocity of 4.2 m/s, off the front of a 25 kg kayak when the kayak is moving forward at a velocity

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Answer:

3.88m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and 2 are the initial velocities

v is the final velocity

Given

m1 = 64kg

u1 = 4.2m/s

m2 = 25kg

u2 = 3.2m/s

Required

Final velocity v

Substitute the given values into the formula

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v = 3.88m/s

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3 years ago

A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled

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3 years ago