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IRINA_888 [86]
4 years ago
14

A team of engineering students is designing a catapult to launch a small ball at A so that it lands in the box. If it is known t

hat the initial velocity vector makes a 32° angle with the horizontal, determine the range of launch speeds for which the ball will land inside the box.
Physics
1 answer:
Murljashka [212]4 years ago
8 0

Answer:

Explanation:

If the initial velocity is U

Then the horizontal component of the velocity is

Ux= Ucosθ

Then the range for a projectile is give as

R=Ux.t

Where t is the time of flight

The time of flight is given as

t=2USinθ/g

Therefore,

R=Ux.t

R=UCosθ.2USinθ/g

R=U^2×2SinθCosθ/g

Then, from trigonometric ratio

2SinθCosθ= Sin2θ

R=U^2Sin2θ/g

Given that θ=32° and g=9.81m/s^2

Then

R=U^2Sin2×32/9.81

R=U^2Sin64/9.81

R=0.0916U^2

Then, range is given by R=0.0916U^2

A=0.0916U^2.

T

The box is at a distance A from the point of projection. Then the range R=A

R=0.0916U^2

A=0.0916U^2

Then,

U^2=A/0.0916

U^2=10.915A

Then the initial velocity should be

U=√10.915A

U=3.3√A

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Answer:

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Answer:

Explanation:

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c₂ = specific heat of aluminum = 0.22 cal/g°C

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