Arocket is launched at an angle of 38o above the horizontal with an initial speed of 78 m/s. it moves for 3 s along its initial line of motion with an acceleration of 12 m/s2 . at this time, its engines fail and the rocket then moves as a projectile. find the maximum height of the rocket and the horizontal range (distance from launch point to landing point).

 y = 428.67 m  and  x all = 1513.68 m

Explanation:

This problem of kinematics can be divided into two parts: a first part when the rockets work a second as a parabolic launch.

Let's do the first part, let's calculate the speed just when the engines turn off

          vf = v₀ + at

          vf = 78 + at

          vf = 78 +12 3

          vf = 114 m / s

This is the speed with which the second part begins vo = 114 m / s with an Angle of 38º

Also at this time a distance is displaced, we calculate the distance traveled (in the direction of the acceleration)

           d = v₀ t + ½ a t²

           d = 78 3 + ½ 12 3²

           d = 288 m

Let's use trigonometry to find the components

          x₀ = d cos 38 = 288 cos 38

          y₀ = d sin38 = 288 sin38

          x₀ = 226.95 m

          y₀ = 177.31 m

Second part

Let's calculate the maximum height, at this point its vertical speed is zero (vfy = 0)

Let's decompose the initial velocity using trigonometry

         vₓ = v₀ cos 38

          = v₀ sin38

         vₓ = 114 cos 38

           = 114 sin38

         vₓ = 89.83 m / s

         = 70.19 m / s

         ² = ² - 2g (y -y₀)

         0 = ² -2g (y -yo)

         y-y₀ = ² / 2g

         y-y₀ = 70.19²/2 9.8

         y = 251.36 + y₀

         y = 251.36 + 177.31

         y = 428.67 m

This is the maximum height from the point where the movement began, that is, the ground.

Now let's calculate the range

         R = vo² sin 2θ / g

         R = 114² sin 2 38 /9.8

         R = 1286.73 m

This is the scope of the parabolic movement, we must add the horizontal distance traveled in the first part

         x all = R + xo

         x all = 1286.73 + 226.95

         x all = 1513.68 m

electrical power

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s