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2 months ago

a stomp rocket takes 1.5 seconds to reach its maximum height what was the initial velocity and what was the maximum height ?

2 answers:

5 0

Here is the solution: 1) Let acceleration be a m/s^2 and velocity after 4 seconds be v m/s. According to v = u + at, we have v = 4a --------(i). For the first phase, applying s = ut + 1/2at^2 results in s1 = 0 + 1/2 x a (4)^2, yielding s1 = 8a ----------(ii). In the second phase, by using s = vt, we have s2 = 4a x (9.1-4), which results in s2 = 20.4a -----------(iii). Adding equations (ii) and (iii) gives us s1+s2 = 8a + 20.4a, leading to 100 = 28.4a, thus a = 3.52 m/s^2. For part 2(a), using v = u - gt gives 0 = u - 9.8 x 1.5, leading to u = 14.7 m/s. For part 2(b), using v^2 = u^2 - 2gh gives 0 = (14.7)^2 - 2 x 9.8 x h, resulting in h = 11.03 m. Thank you for your inquiry.

inna [3.1K]2 months ago

5 0

1) Vf = Vo - gt; Setting Vf = 0 gives Vo = gt, resulting in Vo = 9.8 m/s^2 * 1.5 s = 14.7 m/s. 2) The displacement is calculated as d = Vo*t - gt^2 / 2 = 14.7 m/s * 1.5 - 9.8 m/s^2 * (1.5 s)^2 / 2 = 11.02 m.

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Answer:

Explanation:

a )

Each blade resembles a rod with its axis positioned near one end.

The moment of inertia for one blade is:

= 1/3 x m l²

where m stands for the mass of the blade

l represents the length of each blade.

Total moment of inertia for 3 blades is:

= 3 x $\frac{1}{3}$ x m l²

ml²

2 )

Details provided include:

m = 5500 kg

l = 45 m

Substituting these values produces:

moment of inertia of one blade:

= 1/3 x 5500 x 45 x 45

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Moment of inertia for 3 blades:

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c )

Angular momentum

= I x ω

I denotes the moment of inertia of the turbine

ω symbolizes angular velocity

ω = 2π f

f indicates the rotational frequency of the blades

d )

We have I = 111.375 x 10⁵ kg.m² (Calculated)

f = 11 rpm (revolutions per minute)

= 11 / 60 revolutions per second

ω = 2π f

= 2π x 11 / 60 rad / s

Calculating angular momentum yields

= I x ω

111.375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s

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