Question

A World War II bomber flies horizontally over level terrain with a speed of $\mathbf{275}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$relative to the ground and at an altitude of.$\mathbf{3}\mathbf{.00}\mathit{k}\mathit{m}$ The bombardier releases one bomb. (a) How far does the bomb travel horizontally between its release and its impact on the ground? Ignore the effects of air resistance. (b) The pilot maintains the plane's original course, altitude, and speed through a storm of flak. Where is the plane when the bomb hits the ground? (c) The bomb hits the target seen in the telescopic bombsight at the moment of the bomb's release. At what angle from the vertical was the bombsight set?

Short answer

(a) The bomb travel horizontally between its release and its impacts on the ground is$6.79km$

(b) If The bomb hits the ground, the plane will be vertically above the point where the bomb hits.

(c) Angle from the vertical was the bombsight set${66.2}^{\circ }$

Step-by-step solution

Determine the formula for time

Distance is thus the product of velocity and time. Similarly, you may determine how much time it takes to go a specific distance at a known speed. This time is obtained by dividing the distance by the selected speed.

$t=\frac{d}{s}$

$s=$Speed

$d=$Distance

$t=$Time

Determine the bomb travel horizontally between its release and its impacts on the ground

(a)

The height of the plane is

$h=3.00\text{km}$

$=3000\text{m}$

The horizontal velocity is.$v_h=275\text{m}/\text{s}$ Since there is no air resistance, the only force acting on the shell is gravity, which is vertical. The acceleration due to gravity is.$g=9.81\text{m}/{\text{s}}^2$ So the time taken by the shell to hit the ground is

$t=\sqrt{\frac{2h}{g}}$

$=\sqrt{\frac{2(3000\text{m}/\text{s})}{(9.81\text{m}/\text{s})}}$

$=24.7\text{s}$

Hence, the horizontal distance traveled by the shell is

$R=v_{h}t$

$=(275\text{m}/\text{s})(24.7\text{m}/\text{s})$

$=6792\text{m}$

$=6.79\text{km}$

Hence, the bomb travel horizontally between its release and its impacts on the ground is.$6.79km$

Determine the plane when the bomb hits the ground

(b)

Since the horizontal speed of the shell and the plane is the same, they will travel the same horizontal distance.

Hence, when the bomb hits the ground, the plane will be vertically above the ground where the bomb hits.

Determine the angle from the vertical was the bombsight set

(c)

The horizontal distance traveled by the bomb is $R=6.79\text{km}$and the vertical distance traveled by the bomb is$h=3.00\text{km}$

Hence, the angle with the vertical is

$\theta ={\tan }^{-1}\left(\frac{R}{h}\right)$

$={\tan }^{-1}\left(\frac{6.79\text{km}}{3.00\text{km}}\right)$

$={66.2}^{\circ }$

Hence, the angle from the vertical was the bombsight set${66.2}^{\circ }$