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Talja [164]
3 years ago
9

1.

Physics
1 answer:
8_murik_8 [283]3 years ago
5 0

Answer: 2.61 s

Explanation:

We are given the following data:

s=5 ft is the initial height of the object

v=40 ft/s is the initial height of the object

In addition, the motion of the object is given by:

h=-16t^{2}+ vt +s (1)

Where:

h=0 ft is the final height of the object

t is the time the object is in the air before hitting the ground

Rewritting (1) with the given data:

0=-16t^{2}+ 40t +5 (2)

Solving the quadratic equation with the quadratic formula t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}, where a=-16, b=40, c=5 and choosing the positive value of time:

t=\frac{-40\pm\sqrt{(-40)^{2}-4(-16)(5)}}{2(-16)} (3)

t=2.61 s (4) This is the time

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Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?
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Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

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0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

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Divide both side by – 20

h = – 400 / – 20

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Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

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