Question

.- Un cuerpo es lanzado con una velocidad de 20 m/s, formando un ángulo con la horizontal de 50o, determinar el tiempo de vuelo, el alcance máximo y la altura máxima

Answer 1

- t = 3.12s

- x = 40.19m

- h_max = 11.97m

Step-by-step explanation:

- In order to calculate the flight time of the body, you use the following formula:

`t=(2v_o sin\theta)/(g)` (1)

vo: initial speed = 20m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle with the horizontal of the direction of the motion of the body = 50°

You replace the values of the parameters in the equation (1):

`t=(2(20m/s)sin50\°)/(9.8m/s^2)=3.12s`

The flight time of the body is 3.12s

- The maximum range of the motion is given by:

`x=v_xt=v_ocos\theta t\\\\x=(20m/s)(cos50\°)(3.12s)=40.19m`

where you have used the flight time.

The maximum horizontal distance traveled by the body us 40.19m

- The maximum height is:

`h_(max)=(v_o^2sin^2\theta)/(2g)` (2)

you replace the values of the parameters:

`h_(max)=((20m/s)^2sin^250\°)/(2(9.8m/s^2))=11.97m`

The maximum height is 11.97 m