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riadik2000 [5.3K]
3 years ago
7

A ball is launched into the sky at 54.4 feet per second from a 268.8 meter tall building. The equation for the ball’s height, h,

at time t seconds is h = -3.2t^2 + 54.4t + 268.8. When will the ball strike the ground?
Mathematics
2 answers:
pychu [463]3 years ago
8 0

Answer:

4.42 seconds is the correct answer

Step-by-step explanation:

worty [1.4K]3 years ago
5 0

Answer:

7.41

Step-by-step explanation:

Vertical component:

S = 268.8

U = 0

V =?

A = 9.8

T=?

S = ut + 0.5at^2. u = 0

268.8 = 0.5x9. 8xt^2

T = 7.41 seconds

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3 years ago
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The roof of the Terminal Tower is 709 feet above ground. A window washer dropped a bucket from the edge of the roof. How many se
Masja [62]

Answer:

6.64 seconds

Step-by-step explanation:

Given that:

Distance between roof of the terminal and ground = 709 feet

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To find:

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Solution:

Distance traveled, s = 709 feet

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Acceleration of the bucket will be equal to acceleration due to gravity i.e.

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Let the time taken is t seconds.

Formula for distance is given as:

s = ut+\dfrac{1}{2}at^2

Putting all the values:

709 = 0 (t) + \dfrac{1}{2}(32.1741) t^2\\\Rightarrow 1418 = 32.1741 t^2\\\Rightarrow t^2 = 44.07\\\Rightarrow t \approx \bold{6.64\ sec}

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