A person standing on the top of a building throws a box with a velocity of 7 m/s. If the height of the building is 7 m, determine the horizontal distance the box travels during its flight.

a. 7.36 m
b. 8.36 m
c. 8.86 m         
d. 9.86 m

Initial velocity is v_ox
v_ox = v_1x = 7 m/s
v_oy = 0

Acceleration of a falling particle is
a = -g = - 9.81 m/s²

Let the time taken be T
The vertical distance travelled by the box is given by
y= (v_oy)t - 1/2 g t²        (1)
substituting the values of v_oy and g in (1) we get
7 = 0 - (1/2 ) (9.81) t²
T= 1.1946 s

The horizontal distance travelled by the box is
x = v_1x(T) = 8.36 m.