A dart player throws a dart horizontally at a speed of of 12.4 m/s. The dart hits the board 0.32 meters below the height from which it was thrown. How far away is the player from the board?
h=gt²/2 => t=sqrt(2h/g)
L=vt=v• sqrt(2h/g)
25.17
To find out how far away the player is from the board, we can use the fact that the dart was thrown horizontally.
Since the dart was thrown horizontally, we know that the vertical displacement (change in height) is -0.32 meters. This means the dart hit the board 0.32 meters below the height it was thrown from.
We can use the kinematic equation for vertical motion to solve for the time it takes for the dart to hit the board:
Δy = v0y * t + (1/2) * a * t^2
Since the dart was thrown horizontally, the initial vertical velocity (v0y) is 0 m/s, and the acceleration (a) due to gravity is -9.8 m/s^2. Rearranging the equation:
-0.32 = 0 + (1/2) * (-9.8) * t^2
Simplifying the equation:
0.16 = 4.9 * t^2
Dividing both sides by 4.9:
t^2 = 0.032653
Taking the square root of both sides:
t ≈ 0.1806 seconds
Now, we can use the time to find out how far the player is from the board. Since the dart was thrown horizontally at a speed of 12.4 m/s, the horizontal distance (x) can be found using the equation:
x = v0x * t
where v0x is the initial horizontal velocity. In this case, the horizontal velocity is equal to the speed of the dart, so:
x = 12.4 * 0.1806
x ≈ 2.241 meters
Therefore, the player is approximately 2.241 meters away from the board.
To find the distance the player is from the board, we need to consider the horizontal motion of the dart.
When the dart is thrown horizontally, it will travel in a straight line horizontally and its vertical motion will be affected only by gravity.
In this case, the dart hits the board 0.32 meters below the height from which it was thrown. This means that the vertical displacement (Δy) of the dart is -0.32 meters (negative because it is below the starting height).
The horizontal motion of the dart can be described by the following equation:
Δx = Vx * t
where
Δx is the horizontal displacement
Vx is the horizontal velocity (12.4 m/s in this case)
t is the time taken
We need to find the time taken for the dart to hit the board. To do that, we can use the vertical motion equation:
Δy = Vy * t + (1/2) * g * t^2
where
Δy is the vertical displacement (-0.32 m in this case)
Vy is the vertical velocity (initially 0 since the dart is thrown horizontally)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken
In this case, since there is no initial vertical velocity (Vy = 0), the equation simplifies to:
Δy = (1/2) * g * t^2
Rearranging the equation gives us:
t^2 = (2 * Δy) / g
Substituting the known values, we can calculate the time:
t^2 = (2 * -0.32) / 9.8
Solving for t:
t^2 = -0.064 / 9.8
t^2 ≈ -0.00653
Since the time cannot be negative, we discard the negative value and take the positive square root:
t ≈ 0.0807 s
Now we can substitute the time (t) into the horizontal motion equation to find the horizontal displacement (Δx):
Δx = Vx * t
Δx = 12.4 * 0.0807
Calculating Δx gives us:
Δx ≈ 1.003 meters
Therefore, the player is approximately 1.003 meters away from the board.