A rocket rises vertically, from rest, with an acceleration of$\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$until it runs out of fuel at an altitude of 775 m. After this point, its acceleration is that of gravity, downward.(a) What is the velocity of the rocket when it runs out of fuel?(b) How long does it take to reach this point?(c) What maximum altitude does the rocket reach?(d) How much time (total) does it take to reach the maximum altitude?(e) With what velocity does it strike the earth?(f) How long (total) is it in the air?

Instantaneous velocity is the physical quantity, equivalent to the average velocity taken over an infinitesimally short time interval.

Given data.

The acceleration of the rocket is $a=3.2 \mathrm{m}/{\mathrm{s}}^2$.

The height at which the rocket runs out of fuel is $h=775\mathrm{m}$.

The relation from the equation of motion is given by

$v^2=u^2+2ah$.

Here, v is the final velocity of the rocket, and u is the initial velocity of the rocket, the value of which is zero.

On plugging the values in the above relation,

$\begin{array}{c}{v}^2={\left(0\right)}^2+2\left(3.2 \mathrm{m}/{\mathrm{s}}^2\right)\left(775\mathrm{m}\right)\\ v=70.4\mathrm{m}/\mathrm{s}\end{array}$

Thus, $v=70.4\mathrm{m}/\mathrm{s}$is the required final velocity of the rocket.

The relation from the equation of motion is given by

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

Here, t is the required time taken by the rocket.

On putting the values in the above relation,

$\begin{array}{c}775\mathrm{m}=\left(0\right)t+\frac{1}{2}\left(3.2 \mathrm{m}/{\mathrm{s}}^2\right)t^2\\ t=22.008\mathrm{s}\end{array}$

Thus, $t=22.008\mathrm{s}$is the required time taken by the rocket.

The relation from the equation of motion is given by

$v{\text{'}}^2=v^2+2gh$.

Here, v’ is the velocity at the maximum height, the value of which is zero, g is the gravitational acceleration, and h’ is the altitude.

On plugging the values in the above relation,

$\begin{array}{c}0={\left(70.4\mathrm{m}/\mathrm{s}\right)}^2+2\left(-9.81 \mathrm{m}/{\mathrm{s}}^2\right)h\text{'}\\ h\text{'}=252.6\mathrm{m}\end{array}$

The maximum altitude is calculated as

$\begin{array}{l}H=h+h\text{'}\\ H=\left(775\mathrm{m}\right)+\left(252.6\mathrm{m}\right)\\ H=1027.6\mathrm{m}\end{array}$

Thus, $H=1027.6\mathrm{m}$is the maximum altitude.

The relation from the equation of motion is given by

$v\text{'}=v+gt\text{'}$.

Here, v’ is the velocity at the maximum height, the value of which is zero, g is the gravitational acceleration, and t’ is the time.

On plugging the values in the above relation,

$\begin{array}{c}0=\left(70.4\mathrm{m}/\mathrm{s}\right)+\left(-9.81 \mathrm{m}/{\mathrm{s}}^2\right)t\text{'}\\ t\text{'}=7.17\mathrm{s}\end{array}$

The total time is calculated as

$\begin{array}{l}T=t+t\text{'}\\ T=\left(22.008\mathrm{s}\right)+\left(7.17\mathrm{s}\right)\\ T=29.178\mathrm{s}\end{array}$

Thus, $T=29.178\mathrm{s}$is the total time.

The relation from the equation of motion is given by

${v_0}^2=u^2+2gH$.

On plugging the values in the above relation,

$\begin{array}{c}{v_0}^2={\left(0\right)}^2+2\left(9.81 \mathrm{m}/{\mathrm{s}}^2\right)\left(1027.6\mathrm{m}\right)\\ v_0=141.9\mathrm{m}/\mathrm{s}\end{array}$

Thus, $v_0=141.9\mathrm{m}/\mathrm{s}$is the required velocity.

The relation from the equation of motion is given by

$H=ut\text{'}\text{'}+\frac{1}{2}gt\text{'}{\text{'}}^2$.

On plugging the values in the above relation,

$\begin{array}{c}\left(1027.6\mathrm{m}\right)=0+\frac{1}{2}\left(9.81 \mathrm{m}/{\mathrm{s}}^2\right)t\text{'}{\text{'}}^2\\ t\text{'}\text{'}=14.4\mathrm{s}\end{array}$

The net time can be calculated as

$\begin{array}{l}{T}_{net}=T+t\text{'}\text{'}\\ T_{net}=29.178\mathrm{s}+14.4\mathrm{s}\\ T_{net}=43.578\mathrm{s}\end{array}$

Thus, $T_{net}=43.578\mathrm{s}$is the required total time.