a gun,kept on straight horizontal ground is used to hit a car, traveling along the same road away from the gun with a uniform speed of 72km/h.The car is at a distance of 500m from the gun. when the gun is fired at an angle of 45 degrees with the horizontal. find

(1) the distance of the car from the gun when the shell hit it
(2) the speed of projection of the shell from the gun

Plz answer

To find the distance of the car from the gun when the shell hits it, we need to break down the motion of the shell into horizontal and vertical components.

(1) To find the horizontal distance the car travels, we can use the formula:

distance = speed x time

Given that the car is traveling at a uniform speed of 72 km/h and the time it takes for the shell to hit the car is what we want to find, we can rearrange the formula:

time = distance / speed

Using the given distance of 500 m and converting the car's speed to m/s:

distance = 500 m
speed = 72 km/h = (72 x 1000) m / (60 x 60) s = 20 m/s

Plugging these values into the formula:

time = 500 m / 20 m/s = 25 seconds

Now, to find the vertical distance the shell has traveled in 25 seconds, we can use the formula for vertical displacement in projectile motion:

vertical displacement = (initial vertical velocity x time) + (0.5 x acceleration x time^2)

Given that the shell is fired at an angle of 45 degrees and the initial vertical velocity is 0 m/s, we only need to consider the effect of gravity. The acceleration due to gravity is approximately 9.8 m/s^2.

Using these values:

vertical displacement = (0 x 25 s) + (0.5 x 9.8 m/s^2 x (25 s)^2) = 3,062.5 m

Therefore, the vertical distance the shell has traveled when it hits the car is approximately 3,062.5 m.

Now, to find the distance of the car from the gun when the shell hits it, we can use the Pythagorean theorem:

distance = √((horizontal distance)^2 + (vertical distance)^2)

Plugging in the horizontal distance of 500 m and the vertical distance of 3,062.5 m:

distance = √((500 m)^2 + (3,062.5 m)^2) ≈ 3,106.4 m

Therefore, the distance of the car from the gun when the shell hits it is approximately 3,106.4 m.

(2) To find the speed of projection of the shell from the gun, we can use the formula for projectile motion:

vertical velocity = initial vertical velocity + (acceleration x time)

Given that the initial vertical velocity is 0 m/s and the time it takes for the shell to hit the car is 25 seconds, we only need to consider the effect of gravity. The acceleration due to gravity is approximately 9.8 m/s^2.

Using these values:

vertical velocity = 0 m/s + (9.8 m/s^2 x 25 s) = 245 m/s

Now, to find the speed of projection, we can use the Pythagorean theorem:

speed of projection = √((horizontal velocity)^2 + (vertical velocity)^2)

Since the shell is fired at an angle of 45 degrees, the horizontal and vertical velocities are equal:

horizontal velocity = vertical velocity = 245 m/s

Plugging in this value:

speed of projection = √((245 m/s)^2 + (245 m/s)^2) = √(2 x (245 m/s)^2) ≈ 346.4 m/s

Therefore, the speed of projection of the shell from the gun is approximately 346.4 m/s.