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Maslowich
2 years ago
8

A basketball is tossed upwards with a speed of 5.0 m/s. ​We can ignore air resistance.

Physics
1 answer:
FinnZ [79.3K]2 years ago
4 0

We have that  the maximum height reached by the basketball from its release point is

S=1.3m

From the question we are told

  • A basketball is tossed upwards with a speed of 5.0 m/s. ​We can ignore air resistance.  
  • What is the maximum height reached by the basketball from its release point?

Generally the Newtons equation for Motion  is mathematically given as

V^2=u^2+2as\\\\Therefore\\\\0=25-19.6*S\\\\S=\frac{25}{19.6}\\\\S=1.276

S=1.3m

Therefore

The maximum height reached by the basketball from its release point is

S=1.3m

For more information on this visit

brainly.com/question/23366835

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Answer:

To create a second harmonic the rope must vibrate at the frequency of 3 Hz

Explanation:

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A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
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Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

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Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

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