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Grace [21]
2 years ago
11

7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ba

ll h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.

Physics
1 answer:
Dima020 [189]2 years ago
6 0

\\  \tt \longmapsto \: h =  - 4.8 {t}^{2}  + 8.2t + 1.8 \\ \\  \tt \longmapsto \: h =  - 4.8(1.7) {}^{2}  + 8.2 \times 1.7 + 1.8 \\ \\  \tt \longmapsto \:  - 4.8 \times 2.89 + 1.39 + 1.8 \\ \\  \tt \longmapsto \: 13.8 + 1.39 + 1.8 \\ \\  \tt \longmapsto \: 17.06

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Please help answer question​
nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

\rho is the density of air = 1.225 kg/m³

C is drag coefficient

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C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

So, the drag coefficient is 1.01.

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