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oksano4ka [1.4K]
2 years ago
15

A projectile is thrown with an initial speed of 20 m/s at an angle 30 degrees above

Physics
1 answer:
ZanzabumX [31]2 years ago
4 0
Initial velocity u = 20 m/s

Initial horizontal velocity = 20 cos30° = 20 * 0.866 = 17.32 m/sec.

Initial vertical velocity = 20 sin30° = 20 * 1/2 = 10 m/sec.

time taken t = u/g = 10/10 =1 sec. ( approximating g to 10m/sec^2)

Maximum height h = ut + 1/2 * g * t^2

h = 10 *1 - 1/2 * 10 * 1* 1

h = 10 - 5 = 5 metres.

Total time in air = 2t = 2 seconds.
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A rock is thrown horizontally off a cliff with an initial speed of 3 m/s. The initial position of the rock is 10 meters above th
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Answer:

<em>1.43 s.</em>

Explanation:

Using one of the equations of motion,

S = ut + 1/2gt².......................... Equation 1

Where S = height of the cliff, u = initial velocity, t = time, g = acceleration due to gravity.

<em>Note: When the rock begins to fall from the maximum height, u = 0 m/s, g = positive</em>

<em>Given: S = 10 m, u = 0 m/s</em>

<em>Constant: g = 9.8 m/s²</em>

<em>Substituting these values into equation,</em>

<em>10 = 0(t) + 1/2(9.8)(t²)</em>

<em>10 = 0 + 4.9t²</em>

<em>t² = 10/4.9</em>

<em>t² = 100/49</em>

<em>t = √(100/49)</em>

<em>t = 10/7</em>

<em>t = 1.43 s.</em>

<em>Thus the rock spend 1.43 s in air</em>

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