A dart player throws a dart horizontally at a speed of 13.5 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?
how long does it take the dart to fall 0.32 m?
4.9t^2 = 0.32
with that t, find the distance is
13.5t
To find the distance the player is from the board, we can use the kinematic equation for horizontal motion:
Δx = Vx * t
where:
Δx is the horizontal distance traveled by the dart,
Vx is the horizontal velocity of the dart,
t is the time the dart takes to hit the board.
We are given the horizontal velocity of the dart, Vx = 13.5 m/s.
Now, we need to find the time it takes for the dart to hit the board. To do this, we can use the kinematic equation for vertical motion:
Δy = Vy * t + (1/2) * g * t^2
where:
Δy is the vertical distance traveled by the dart,
Vy is the vertical velocity of the dart,
g is the acceleration due to gravity (approximately 9.8 m/s²).
We are given that Δy = -0.32 m (negative because the dart hits below the initial height) and Vy = 0 m/s (since the dart was thrown horizontally).
Plugging in the values, we have:
-0.32 = 0 * t + (1/2) * 9.8 * t^2
Simplifying the equation:
-0.32 = (4.9 * t^2)
Dividing both sides by 4.9:
-0.32 / 4.9 = t^2
-0.06531 = t^2
Taking the square root of both sides:
t ≈ √(-0.06531)
Since we cannot take the square root of a negative number, it means that there is no real solution for t. This means that the dart will never hit the board.
Therefore, there is no distance to be calculated for the player from the board because the dart will not reach the target.