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?

1 Answer
Mar 3, 2016

Answer:

#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#