Question

A projectile is shot from the ground at a velocity of `4 m/s` at an angle of `pi/6`. How long will it take for the projectile to land?

Answer 1

`4* sqrt(3)/5 m`

Explanation:

Let's start with Newton's equation `**F** = m* **a**`. As the problem is bi-dimensional, the previous equation can be split in the horizontal and vertical components

`F_x = m*a_x` for the horizontal component and

`F_y = m*a_y` for the vertical component.

The horizontal acceleration is null while the vertical acceleration is equal to gravity

`a_x = 0` ; `a_y =-g`

and using the definition of acceleration

`a_x = (d^2x)/dt^2 ; a_y = (d^2y)/dt^2`

and integrating both part of equations

`x= v_(x0)*t ; y= v_(yo)*t - g*t^2/2 (1)`

The projectile will land when `y=0 =>` introducing it in (1)

`0= v_(yo)*t_l - g*t_l^2/2 ; t_l = 2*v_(yo)/g`

and replacing in the equation or `x`

`x_l= 2*v_(x0)*v_(y0)/g`

The components of shot speed are

`v_(x0)= v_0 *cos(pi/6)`

`v_(y0)= v_0 *sin(pi/6)`

thus

`x_l= 2*v_0^2/g*cos(pi/6)*sin(pi/6)`

and replacing by the figures given in the problem

`x_l= 2*16/10 m*cos(pi/6)*sin(pi/6) = 16/5m*1/2*sqrt(3)/2 =` `=4* sqrt(3)/5 m`