A dart is thrown horizontally with an initial speed of 10 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 ? (b) How far away from the dart board is the dart released?

Question

mani_jha · Answer

Though the dart is directed towards point P, due to the force of gravity it will be pulled downwards and will hit a point directly below P. Which means that if you play darts in a space station, and neglect air ressistance, you will hit exactly the point you aimed for!
Due to the force of gravity, the dart will experience an acceleration downwards. So, what distance will it cover downwards in 0.19 seconds(Hint: Use the equations of motion under constant acceleration.)
Do you know those equations?

anonymous · Answer

ah yes

anonymous · Answer

so you use the equation for horizontal motion x-x0=v0xt to find its xposition

then use the other equation
 y-y0=v0yt - 1/2gt^2

mani_jha · Answer

Good, but the equations are:
x-x0=v0t
y-y0=vot-1/2gt^2.
For the first part, use the second equation and find (y-y0)
For the first part, use the first equation and find (x-x0)