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ruslelena [56]
3 years ago
5

Calculus application

Mathematics
1 answer:
Ugo [173]3 years ago
8 0

Answer:

49m/s

59.07 m/s

Step-by-step explanation:

Given that :

Distance (s) = 178 m

Acceleration due to gravity (a) = g(downward) = 9.8m/s²

Velocity (V) after 5 seconds ;

The initial velocity (u) = 0

Using the relation :

v = u + at

Where ; t = Time = 5 seconds ; a = 9.8m/s²

v = 0 + 9.8(5)

v = 0 + 49

V = 49 m/s

Hence, velocity after 5 seconds = 49m/s

b) How fast is the ball traveling when it hits the ground?

V² = u² + 2as

Where s = height = 178m

V² = 0 + 2(9.8)(178)

V² = 0 + 3488.8

V² = 3488.8

V = √3488.8

V = 59.07 m/s

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Answer:

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Step-by-step explanation:

According to the scenario, computation of the given data are as follows,

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a^{2} + b^{2} = c^{2}

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In 2001 the mint indoor pool vault record was 20 and 1/6 the women’s record for the indoor pool vault was 15 and 5/12 FT what is
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Answer:

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Step-by-step explanation:

Height of the indoor pool vault record for men = 20 and 1/6 ft =\frac{121}{6} ft

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The combined height of the two records :

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