1 A projectile on Earth is launched with a velocity of 200 m

1.) A projectile on Earth is launched with a velocity of 20.0 m/s and lands in a bucket 6.00 meters away and 2.00 meters below the launch plane. What is the largest angle of trajectory in (deg)?

2.) If a projectile is shot at a velocity of 90.0 m/s and lands 800.0 meters on the same plane. What is the highest possible launch angle (deg)?

3.) A pumpkin is launched at 100.0 m/s at an angle of 30.0º to what height in meters (m)?

Solution

the kinematic equation,

     y = 1/2 a t^2 + v(y)*t + y(o)
    -2 = 1/2 * (-9.8) (4.18)^2 + v(y)*4.18 + (-2)
but, y =y(o) = -2

So, the above equation becomes as follows:

   0 = 1/2 * (-9.8) * (4.18)^2 + v(y) * 4.18
0 = -85.6 + v(y) * 4.18

thus, the velocity is,
v(y) = 85.6/4.18

     = 20.4785 m/s

thus, the angle,

@ = tan^-1 ( v(y) / v(x) )

      = tan^-1 ( 20.4785 / 1.435)

      = 86^o