| Physics High School |
| # h2phyt1 -
Projectile motion |
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A stone is thrown from the edge of a vertical cliff with a velocity of 50 m•s
-1
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at an angle tan-1
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above the horizontal. The stone strikes the sea at |
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a point 240 m from the foot of the cliff.
Find the time for which the stone
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is in the air and the height of the cliff.
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Strategy - the rule "Must have" of Mr.Zhang ®
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1. Convert tan-1
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to be sinθ and cosθ |
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2. Formulae : Sy = vyt + |
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gt2
| (vertical) → (1) |
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S x= vxt (horizontal) → (2) |
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Horizontal motion (time) = Vertical motion (time) |
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Solution |
Compute the time to find the height |
| ∵ tan-1 |
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| Therefore,tanθ= |
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,
sinθ= |
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,
cosθ= |
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Horizontal motion from equation (2);
| S x | = | vxt |
| S x | = | vxcosθt |
240 m | = | 50 m•s-1 |
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t |
| t | = | 5 s |
We bring the horizontal time to calculate the vertical
motion to get the height value.
| Sy = 50 m•s-1sinθ(5) + |
| (-9.8 m•s-2)(52)
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Gravitational against downward direction of the earth,
therefore, we take the minus sign. |
| -h = 50 m•s-1
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(5) + |
| (-9.8 m•s-2)(52)
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Displacement direction required is lower the level of the cliff.
Hence we take the minus sign. |
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Ans.
1. The stone is in the air for 5 second.
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2. The height of the cliff 52.5 m.
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