Question

You fire a projectile 35° above horizontal with an initial velocity of 200m/s. It lands in a valley 300m below the launch point. What is the time of flight of the projectile, and what is the range of projectile? ​

Answer 1

Answer:

See below

Explanation:

Vertical component of initial velocity  =  200 sin 35° = 114.72 m/s

then use position formula      a = 9.81 m/s^2

  0  = 300 + 114.72 t  -  1/2 (9.81)(t^2)

            use quadratic formula with a = - 4.905    b = 114.72   c = 300

              to find t = 25.76 seconds

To find the range

   ( horizontal distance the projectile lands from launch point)

Horizontal component of initial velocity    200 cos 35 = 163.83 m/s

 ( it flies horizontally at this speed for  25.76 seconds <==found above)

    163.83  m/s  *  25.76 s = 4220.3 meters