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3 years ago

How do you calculate time from launch to explosion in a firework

Physics

2 answers:

xenn [34]3 years ago

8 0

Answer:

Explanation:

If a rocket (firework) is launched

with an initial velocity of 39.2 meters per second at a height of 1.6 meters above

the ground, the equation h = -4.9t2

+ 39.2t + 1.6 represents the rocket’s height h in

meters after t seconds. The rocket will explode at approximately the highest point.

Fire official require the fireworks to explode at a height greater than 76 feet in

order for debris to land in Lake Michigan and not on the festival sight. The event

planners assured the fire officials that the fireworks would reach a height greater

than 76 feet

Sholpan [36]3 years ago

4 0

Answer:

If a rocket (firework) is launched with an initial velocity of 39.2 meters per second at a height of 1.6 meters above the ground, the equation h = -4.9t2 + 39.2t + 1.6 represents the rocket's height h in meters after t seconds. The rocket will explode at approximately the highest point.

Explanation:

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How does density help scientists?

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3 years ago

Read 2 more answers

An arrow of 43 g moving at 84 m/s to the right, strikes an apple at rest. The arrow sticks to the apple and both travel at 16.8

Aloiza [94]

Answer:

The mass of the apple is 0.172 kg (172 g)

Explanation:

The Law Of Conservation Of Linear Momentum

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of both momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

We are given the mass of an arrow m1=43 g = 0.043 kg traveling at v1=84 m/s to the right (positive direction). It strikes an apple of unknown mass m2 originally at rest (v2=0). The common speed after they collide is v'=16.8 m/s.

We need to solve the last equation for m2:

$m_2v_2-m_2v'=m_1v'-m_1v_1$

Factoring m2 and m1:

$m_2(v_2-v')=m_1(v'-v_1)$

Solving:

$\displaystyle m_2=\frac{m_1(v'-v_1)}{v_2-v'}$

Substituting:

$\displaystyle m_2=\frac{0.043(16.8-84)}{0-16.8}$

$\displaystyle m_2=\frac{-2.8896}{-16.8}$

$\displaystyle m_2=0.172\ kg$

The mass of the apple is 0.172 kg (172 g)

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