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Sati [7]
2 years ago
11

A child jumps from a moving sled with a speed of 2.2 m/s and in the

Physics
1 answer:
butalik [34]2 years ago
4 0

The initial velocity of the sled will be 7.34 m/sec. V is the initial velocity of the sled.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁)  mass of child = 38 kg

(u₁) is the initial velocity child = 2.2 m/s

(m₂) is the mass of sled = 68 kg

(u₂) is the initial velocity of sled = ?

(v) is the velocity after collision = 5.5 m/s

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

\rm  m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\(38 \times  2.2) + (68 \times V) = 5.5 \times (38+68) \\\\ 83.6 + 68V = 5.5 \times 106 \\\\  83.6 + 68V =583 \\\\ \rm 68\ V= 583 -83.6 \\\\ \rm 68 \ V= 499.4\\\\ V= 7.34 \ m/sec

Hence,the initial velocity of the sled will be 7.34 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
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b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

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v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
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