Question

Determina la velocidad y la distancia total recorrida por un móvil que se desplaza con trayectoria parabólica, tomando en cuenta que su velocidad inicial es de 50 m/s y el ángulo de lanzamiento es de 81º. Ayuda porfavor

Answer 1

First of all, in a parabolic motion, there are two coordinates, horizontal and vertical. That means we need to find the distances and velocities of both coordinates.

We know that
`v_(0)=50 m/s` and
`\theta = 81\°`. So, we can already find the initial velocity for both coordinates.

`v_{0_(y) } =v_(0) sin(\theta) = 50 (sin81\° ) \approx 49.4 m /s\\v_{0_(x) } =v_(0) cos(\theta) = 50 (cos81\° ) \approx 7.8 m/s`

Which makes totally sense, beacuse the angle of shooting is near 90°, as a result, the projectile will cover more vertical distance than horizontal distance.

Now, we know that a parabolic motion divides into two congruent part, when going up and when going down. Also, when the projectile reaches the maximum height, the vertical velocity there is null. So, we use this information to find the maximum time.

`v_(y)=v_{0_(y) } +gt\\ 0= 49.4 + (-9.8)t\\t_(max) =(-49.4)/(-9.8) \approx 5.04`

So, the projectile takes 5.04 seconds to reach the maximum height. The total time of the whole parabolic motion would be double (due to the symmetry)

`t=2(5.04) =10.08`

Therefore, the projectile travels 10.08 seconds in total.

Now, we can find the maximum height reached by the projectile.

`v_(f)=v_{0_(y) }+2gh\\ 0= 49.4+2(-9.8)h\\h=(-49.4)/(-19.6) \approx 2.5`

So, the maximum height reached is 2.5 meters. The total vertical distance of the movement is 5 meters (due to the symmetry).

Now, we just need to find the horizontal distance.

`x=v_{0_(x) } t= 7.8(10.08)=78.62`

The total horizontal distance traveled is 78.62.

At last, the final speed when the projectile reaches the ground after being launched is the same as the initial, beacuse the movement is symmetric.