Motion in Two Dimensions – Projectile Motion

Example

From a tower (height : 50 m) an object is projected  horizontally with a velocity of 20 ms-1

Projectile motion

Ignoring the air resistance:

a.  Determine at what place the object hits the ground, ignoring the air resistance ?

b.  With what velocity does that happen ?

c.   With what angle the object hits the ground ?

Solution:

The directions to the right and upward have been chosen as positive.

a. In the vertical (y)  direction the movement has a constant acceleration due the gravity.

ay = g = – 9.81 m s-2.  Initial velocity in y direction = 0 ms-1

       Δ Y =  ½ g t2    – 50 = ½ -9.81 t2      t2 =  10.20     t = 3.2 s

      In the horizontal (x)direction the object has a constant velocity

     ΔX = vx t         x = 20 x 3.2 = 64 m

      The object hits the ground at 64 m from the base of the tower.

b. The velocity in x – direction remains  constant.

vx van 20 m/s

        vy = g t = -9.81 x 3.2 = -31.39 m/s  ( minus because the direction of the velocity is downward)

        The magnitude of the final velocity is to be determined using the Pythagorean theorem

        v2 = vx2 + vy2        v2 = 202 + 31.392       v2 = 1385.33    v = 37.2 m/s

c. tan α = vy/vx = 31.39/20 = 1.5695      α = 57.5 o

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