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tigry1 [53]
3 years ago
14

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 . Th

e acceleration period lasts for time 5.00 s until the fuel is exhausted. After that, the rocket is in free fall. Find the maximum height ymax reached by the rocket. Ignore air resistance and assume a constant free-fall acceleration equal to 9.80 m/s2 .
Physics
1 answer:
gtnhenbr [62]3 years ago
8 0

Hi there!

We can begin by using the following formulas:

Δd = vit + 1/2at²

vf² = vi² + 2ad

v = at

We can break this problem into 2 parts:

Distance traveled with acceleration:

Δd = vi(t) + 1/2at²

The rocket starts from rest so disregard the "vi(t)":

Δd = 1/2at²

Plug in the given acceleration and time:

Δd = 1/2(29.4)(5)²

Δd = 367.5 m

Distance traveled after thrust stops:

Find the displacement up to the top of the trajectory using the equation:

vf² = vi² + 2ad, where vf = 0 m/s

Calculate the initial velocity using the equation v = at:

v = (29.4)(5) = 147 m/s

Use the above equation and plug in the given values:

0² = (147)² + 2(-9.8)d

Solve for d:

0 = 21,609 - 19.6d

19.6d = 21,609

d = 1,102.5 m

Add the two displacements:

d1 + d2 = 1,470 m

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Answer:

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Answer:

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

Explanation:

Hi there!

The average velocity is calculated as the displacement of the object divided by the time it takes the object to do that displacement.

The displacement is calculated as the distance between the final position of the object and the initial position. <u>In this problem</u>, the displacement is equal to the traveled distance because the object travels only in one direction:

a.v = Δx/t

Where:

a.v = average velocity.

Δx = displacement = final position - initial position

t = time

So, let's find the distance traveled while the object was accelerating. For that, we will use this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

In this case, since the object is initially at rest, v0 = 0. If we place the origin of the frame of reference at the point where the object starts moving, then x0 = 0. So, the equation of the position of the object after a time t will be:

x = 1/2 · a · t²

x = 1/2 · 1.90 m/s² · (6.60 s)²

x = 41.4 m

The object traveled 41.4 m during the first 6.60 s.

Now, let's find the rest of the traveled distance.

When the velocity is constant, a = 0. Then, the equation of position will be:

x = x0 + v · t

Let's place now the origin of the frame of reference at the point where the object starts traveling at constant velocity so that x0 = 0:

x = v · t

The velocity reached by the object during the acceleration phase is calculated as follows:

v = v0 + a · t   (v0 = 0 because the object started from rest)

v = 1.90 m/s² · 6.60 s

v = 12.5 m/s

Then, the distance traveled by the object at a constant velocity will be:

x = 12.5 m/s · 8.50 s

x = 106 m

The total traveled distance in 15.1 s is (106 m + 41.4 m) 147 m.

Then the displacement will be:

Δx = final position - initial position

Δx = 147 m - 0 = 147 m

and the average velocity will be:

a.v = Δx/t

a.v = 147 m / 15.1 s

a.v = 9.74 m/s

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

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We need to find the change in momentum of the tennis ball. It is equal to the final momentum minus initial momentum

\Delta p=m(v-u)\\\\=0.06\times (26-20)\\\\=0.36\ kg-m/s

So, the change in momentum of the ball is 0.36 kg-m/s.

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