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3 years ago

A ball is thrown vertically upwards with a velocity of 9.8 m/s. Its velocity after 1 second will be

Physics

1 answer:

Reil [10]3 years ago

5 0

$\LARGE{ \underline{\underline{ \purple{ \bf{Required \: answer:}}}}}$

GiveN:

Initial velocity = 9.8 m/s²
Accleration due to gravity = -9.8 m/s²
Time taken = 1 s

To FinD:

Final velocity of the ball?

Step-by-step Explanation:

Using the first Equation of motion,

⇒ v = u + gt

⇒ v = 9.8 + -9.8(1)

⇒ v = 0 m/s

The final velocity is hence 0 m/s.

While solving questions of under gravity motions using equations of motion, remember the sign convection to avoid mistakes.
You can consider positive above the ground and negative for towards it.

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A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p

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Answer:

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

Explanation:

The knowable variables are

$d_{t}=25m$

$t_{1}=1.3 s$

$t_{2}=3.9 s$

$t_{3}=5.5 s$

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a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

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Brilliant_brown [7]

The question isn't clear enough, I think it ask us to calculate the linear speed of a point at the edge of the DVD.
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\\
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$
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