A battleship simultaneously fires two shells toward two identical enemy ships. One shell hits ship A, which is close by, and the other hits ship B, which is farther away. The two shells are fired at the same speed. Assume that air resistance is negligible and that the magnitude of the acceleration due to gravity is g.1. What shape is the trajectory (graph of y vs. x) of the shells?a. straight lineb. parabolac. hyperbolad. the shape cannot be determined.2. For two shells fired at the same speed which statement about the horizontal distance traveled is correct?A. The shell fired at a larger angle with respect to the horizontal lands farther away. B. The shell fired at an angle closest to 45 degrees lands farther away. C. The shell fired at a smaller angle with respect to the horizontal lands farther away. D. The lighter shell lands farther away. Now, consider for the remaining parts of the question below that both shells are fired at an angle greater than 45 degrees with respect to the horizontal. Remember that enemy ship A is closer than enemy ship B.3. Which shell is fired at the larger angle?A. AB. BC. Both shells are fired at the same angle.4. Which shell is launched with a greater vertical velocity, vy? A. AB. BC. Both shells are launched with the same vertical velocity.5. Which shell is launched with a greater horizontal velocity, vx?A. AB. BC. Both shells are launched with the same horizontal velocity.6. Which shell reaches the greater maximum height?A. AB. BC. Both shells reach the same maximum height.7. Which shell has the longest travel time (time elapsed between being fired and hitting the enemy ship)?A. AB. BC. Both shells have the same travel time.

1) the correct answer is B ,  2)  θ= 45º,

3) the projectile A must have a greater angle ,

4A)   vₓ (B)> vₓ (A) ,  

5B)  The maximum height is reached by the projectile with the highest speed, from the front this is projectile B

7) if the vertical velocity of both projectiles is the same time is the same

therefore the two projectiles have equal flight times

Explanation:

In this long exercise you are asked to answer a series of questions about the launching of projectiles

1) Path type

since they indicate that the air resistance is negligible the horizontal displacement is

       x = v₀ₓ t

On the vertical axis, a relationship acts, which is gravity, therefore its displacement is

      y = t - ½ g t²

let's substitute to eliminate time

     y = v_{oy} (x / v₀ₓ) - ½ g (x / v₀ₓ)²

    y = (tan θ) x - (g / v₀ₓ²) x²

we can see that this is the equation of a parabola

therefore the correct answer is B

2) From the missile launch expressions the range is

           R = x = v₀² sin² 2θ / g

for the range to be maximum sin 2θ = 1, this occurs for

          2θ = 90

         θ= 45º

the correct answer is "The projectile fired at an angle close to 45º lands farther"

In response D implies that the projectile has been divided into two parts, in this case it is true that the lighter part must go further.

3) From the expression of scope we can see that as the angle increases the square sine becomes smaller, for example without 180 = 0

therefore to reach the nearest ship A must launch with a greater angle, therefore the projectile A must have a greater angle

Respect the questions are a bit strange.

* If the projectiles are fired at the same angle and in the statement they indicate that the velocities are equal, the two must reach the same height 4

4A. If the vertical speed is equal, the projectile that must reach the furthest ship (B) must have more horizontal speed

          vₓ (B)> vₓ (A)

5B. Both are launched with the same horizontal speed.

which is higher

            ² =   v_{oy}² - 2 g y

at maximum height vertical velocity is zero (v_{y} = 0)

            0 = v_{oy}² - 2 g y

            y = v_{oy}² / 2g

the maximum height is reached by the projectile with the highest speed, from the front this is projectile B

7) they both reach the same height

As the value of the acceleration due to gravity is constant, the time to go up is the same time to go down, therefore if the two projectiles reach the maximum height, the time used is

              = v_{oy} - g t

              0 = v_{oy} - gt

              t = v_{oy} / g

if the vertical velocity of both projectiles is the same time is the same

therefore the two projectiles have equal flight times

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.