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3 years ago

5

A catapult used by medieval armies hurls a stone of mass 32.0 kg with a velocity of 50.0 m/s at a 30.0 degree angle above the ho

rizontal. What is the horizontal distance traveled when the stone returns to its original height? Ignore air resistance. Check your calculator deg/rad mode.

1 answer:

NISA [10]3 years ago

5 0

Answer:

The horizontal distance traveled when the stone returns to its original height = 220.81 m

Explanation:

Considering vertical motion of catapult:-

At maximum height,

Initial velocity, u = 50 sin30 = 25 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v = u + at

Substituting

v = u + at

0 = 25 - 9.81 x t

t = 2.55 s

Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s

Considering horizontal motion of catapult:-

Initial velocity, u = 50 cos30 = 43.30 m/s

Acceleration , a = 0 m/s²

Time, t = 5.10 s

We have equation of motion s= ut + 0.5 at²

Substituting

s= ut + 0.5 at²

s = 43.30 x 5.10 + 0.5 x 0 x 5.10²

s = 220.81 m

The horizontal distance traveled when the stone returns to its original height = 220.81 m

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Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

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The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

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$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

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