Question

A dart is thrown horizontally with an initial speed of 16 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.19 s later. (a) What is the distance PQ? (b) How far away from the dart board is the dart released?

Answer 1

a)
y = -1/2 g t^2 = 0.5*9.81*.19^2=0.177 m
b)
x = vt = 17*.19=3.23 m

Answer 2

Since the dart is experiencing a downward net force due to gravity, you can use dist = 1/2 a t2. The horizontal velocity is irrelevant because you are only dealing with the vertical movement.
dist = 1/2(9.81 m/s2)(0.23 s)(0.23 s)

Answer 3

Inital Vertical Velocity , u = m/sec


Acceleration due to Gravity , g = 9.8 m/sec^2

Therefore

S = ut + 0.5*g*t^2

S = 0 + 0.5*g*t^2 = 0.5*9.8*0.19^2 = 0.177 m

Therefore

PQ = 0.177 m

Horizontal Distance = Horizontal Speed*Time

= 10*0.19 = 1.9 m