A dart player throws a dart horizontally at 12.4 m/s. The dart hits the board 0.32 m. below the height from..

Question

A dart player throws a dart horizontally at 12.4 m/s. The dart hits the board 0.32 m. below the height from which it was thrown. How far away is the player from the board?

Please help me with this problem.
I'm a confused.

Answer 1

the thing this problem is concerned is the vertical component of the projectile.

since the dart is thrown horizontally, the initial vertical velocity is zero.

the vertical velocity of the dart just before it hits the dart board is

v^2 = u^2 + 2gs
v^2 = 0 + 2(9.8)(0.32)
v^2 = 6.272
v = 2.50 m/s

the time it took the dart to hit the dart board is given by

v = u + gt
2.50 = 0 + 9.8t
t = 0.255 sec


now, we have the time of travel of the dart, we can use the formula d = vt to find the horizontal distance
(the horizontal velocity remained constant throughout the flight)

d = vt
d = 12.4m/s * 0.255s
d = 3.169 m


therefore the dart player is standing 3.169 meters away from the board...

there ya go...

Answer 2

3.17 m

first, compute the flight time by using the drop of the dart due to gravity
d=1/2*a*t^2
t=sqrt(2*d/a)

once you know the flight time, compute the distance using the horizontal velocity:
t*v=d

12.4*sqrt(2*.32/9.8)=d

j

Answer 3

you have to derivate the formula for gravatational accelleration to find time it took the dart to drop .32 m,, that is your time to apply to the speed to get your distance. I'm too lazy to look it up.