**Example**

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

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.

a_{y}_{ }= g = – 9.81 m s^{-2}. Initial velocity in y direction = 0 ms^{-1}

Δ Y = ½ g t^{2 } – 50 = ½ -9.81 t^{2 } t^{2} = 10.20 t = 3.2 s

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

ΔX = v_{x} 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.

v_{x} van 20 m/s

v_{y} = 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

v^{2} = v_{x}^{2 }+ v_{y}^{2} v^{2} = 20^{2} + 31.39^{2} v^{2} = 1385.33 v = 37.2 m/s

**c. **tan α = v_{y}/v_{x }= 31.39/20 = 1.5695 α = 57.5^{ o}