Chapter 2, Problem 75GP

Solution

a.

The velocity of the rocket, when it runs out of the fuel.

  $v=87.64$ m/sec.

Given:

Acceleration of rocket starting from rest is 3.2 m/sec²

Rocket runs out of fuel at an altitude of 1200 m.

Formula used:

  $v^2=u^2+2ah$ here v is the final velocity, u is initial velocity , a is acceleration and h is height

Calculation:

For the given situation, u = 0, h = 1200 m, a = 3.2m/sec²

  $\begin{array}{c}{v}^2=0+\left(2\times 3.2\times 1200\right)\\ v^2=7680\\ v=87.635\\ \approx 87.64\text{m}/\text{s}\end{array}$

b.

Time taken by the rocket to reach the given height.

  $t=27.39\sec$

Given:

Acceleration of rocket starting from rest is 3.2 m/sec²

Rocket runs out of fuel at an altitude of 1200 m.

Formula used:

  $h=ut+\frac{1}{2}a{t}^2$

Here h is height, u is initial velocity, a is acceleration and t is time Calculation:

From $h=ut+\frac{1}{2}a{t}^2$

  $1200=0+0.5\times 3.2\times t^2$

  $\begin{array}{c}{t}^2=\frac{1200}{1.6}\\ =\frac{3000}{4}\\ =750\\ t=27.386\\ =27.39\text{sec}\end{array}$

c.

The maximum altitude does the rocket reach.

Maximum altitude = 1591.84m

Given:

Acceleration of rocket starting from rest is 3.2 m/sec²

Rocket runs out of fuel at an altitude of 1200 m.

Formula used:

The equation of motion is,

  $v^2=u^2-2gh$

Calculation:

After 1200 m height it will go upwards until its velocity becomes zero

Let it goes up to h¹m height after 1200m.

  $v^2=u^2-2g{h}^1$

  $0={(87.64)}^2-2\times 9.8\times h^1$

  $h^1={\frac{(87.64)}{2\times 9.8}}^2$

  $h^1=391.84\text{m}$ m

The maximum altitude that the rocket reach $=1200+391.84=1591.84$ m

d.

Time taken by the rocket to reach the maximum height.

  $t=36.33\sec$

Given:

Acceleration of rocket starting from rest is 3.2 m/sec²

Rocket runs out of fuel at an altitude of 1200 m.

Formula used:

The equation of motion is,

  $v=u-gt$

Calculation:

Let t¹is the time to reach from 1200m to the height when velocity becomes zero.So,

  $v=u-g{t}^1$

  $0=87.64-9.8t^1$

  $t^1=\frac{87.64}{9.8}$

  $t^1=8.94$ sec

Total time is,

  $\begin{array}{c}T=27.39+8.94 \\ =36.33\text{sec}\end{array}$

e.

The velocity at the instant of striking the Earth.

  $v=\sqrt{31200}=176.64$ m/sec

Given:

Acceleration of rocket starting from rest is 3.2 m/sec²

Rocket runs out of fuel at an altitude of 1200 m.

Formula used:

  $v^2=u^2+2gh$ , here v is final velocity, u is initial velocity , h is height and g is accelerationCalculation:

From previous parts, $h=1591.84\text{m}$

Substituting the values,

  $v^2=0+2\times 9.8\times 1591.84=31200$

  $v=\sqrt{31200}=176.64\text{m}/\text{sec}$

f.

Total time for which the rocket is in the air.

Total time = 54.35sec

Given:

Acceleration of rocket starting from rest is 3.2 m/sec²

Rocket runs out of fuel at an altitude of 1200 m.

Formula used:

The equation of motion is,

  $h=ut+\frac{1}{2}g{t}^2$

Calculation:

Let the time taken to reach the ground from maximum altitude is $t"$

  $h=ut+\frac{1}{2}gt{"}^2$

  $1591.84=0+\frac{1}{2}\times 9.8\times t{"}^2$

  $t{"}^2=\frac{1591.84}{4.9}$

  $t"=18.02\sec$ sec

Total time is,

  $\begin{array}{c}{T}_t=27.39+8.94+18.02\\ =54.35\sec \end{array}$