An us bomber is flying horizontally at 300 mph at an altitude of 610 m. its target is an iraqi oil tanker crusing 25kph in the same direction and same vertical plane. what horizontal distance behind the tanker must the pilot observe before he releases the shell to score a direct hit

=1419.19 meters.

Explanation:

The time it takes for the shell to drop to the tanker from the height, H =1/2gt²

610m=1/2×9.8×t²

t²=(610m×2)/9.8m/s²

t²=124.49s²

t=11.16 s

Therefore, it takes 11.16 seconds for a free fall from a height of 610m

Range= Initial velocity×time taken to hit the tanker.

R=v₁t

Lets change 300 mph to kph.

=300×1.60934 =482.802 kph

Relative velocity=482.802 kph-25 kph

=457.802 kph

Lets change 11.16 seconds to hours.

=11.16/(3600)

=0.0031 hours.

R=v₁t

=457.802 kph × 0.0031 hours.

=1.41918 km

=1.41919 km × 1000m/km

=1419.19 meters.

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

force tends to be low and long diustance masses are high