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Ierofanga [76]
2 years ago
14

'Suppose your hand moves upward by 0.50m while you are throwing the ball. The ball leaves your hand with an upward velocity of 2

0.0 m/s. Find the magnitude of the force (assumed constant) that your hand exerts on the ball. Ignore air resistance.' Now my question is not 'What is the magnitude' but rather: why did I get (roughly) the same answer using F=ma when you were supposed to use the total mechanical energy (W+K1+U1=K2+U2). So I'm more confused about how the 2 formulas are 'related', what the force actually represents in both, when to use what and if there is an actual difference.

Physics
1 answer:
REY [17]2 years ago
6 0

The formula F = ma is used when the statement of second law of motion is applied to the problem. If the statement is applied to the problem, The fomula will be used. If not, use the formula W +  K_{1} +  U_{2} =  K_{2} +  U_{2}

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PLEASE HELP
Sergeu [11.5K]

The vertical component of the initial velocity is v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is v_0_x = \frac{x}{t}

The horizontal displacement when the object reaches maximum height is X = \frac{xy}{gt^2} + \frac{x}{2}

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is calculated as;

x = v_0_xt\\\\v_0_x = \frac{x}{t}

The time to reach to the maximum height is calculated as;

T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T =  \frac{1}{g}  (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t

The horizontal displacement when the object reaches maximum height is calculated as;

X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}

Learn more here: brainly.com/question/20689870

8 0
3 years ago
Remember KE=M*V2/2…
 A model airplane moves twice as fast as another identical model airplane. Compared with the kinetic energy
pishuonlain [190]

Answer: 4 times as much

Explanation:

4 0
2 years ago
A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?
Anon25 [30]

Answer:

57N

Explanation:

F_F = \mu F_N = \mu mg = 0.11 \times 53 kg \times 9,8 \frac{m}{s^{2} }

6 0
2 years ago
Read 2 more answers
If you can answer all of this then your a legend (I'm giving you all my points)
navik [9.2K]

Answer:

1. B

2. B

3. D

4. A

Hope this was correct! A lot of the answers are already in the article itself and the wording is just different. I suggest now that for information retainment, you read the article again with the correct points in mind and see if you can spot the points where the answers are stated!

3 0
2 years ago
How much time does it take for a rattlesnake to strike its victim if it is strikes at a speed of 1.25 m/s and the victim is exac
Marianna [84]

Answer:

4 seconds

Explanation:

5/1.25=4

6 0
2 years ago
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