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barxatty [35]
3 years ago
6

A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. What is the vertical compon

ent of the cannonball’s velocity? What is the horizontal component of the cannonball’s velocity?
473.8 m/s; 473.8 m/s
-525.2 m/s; 435.5 m/s
0 m/s; 670 m/s
-378 m/s; 378 m/s
Physics
1 answer:
elixir [45]3 years ago
6 0

Answer:

<h3>473.8 m/s; 473.8 m/s</h3>

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity  = 670 (0.7071)

Vertical component of the cannonball’s velocity  = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

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A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if t
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Explanation:

From the law of conservation of momentum:

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Therefore,

(97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

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s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)

<u>s₁ = 0.022 m</u>

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