Question

A dart is thrown horizontally at a speed of 12 m/s at the bull's-eye of a dartboard 2.7 m away. How far below the intended target does the dart hit

Answer 1

To determine the distance the dart falls below the intended target, we need to use the equations of motion for projectile motion. The vertical motion is affected by gravity and can be described by the equation:

y = vi_y * t + 1/2 * a * t^2

where
vi_y = initial vertical velocity (0 m/s, since the dart is thrown horizontally)
a = acceleration due to gravity (9.8 m/s^2, downward)
t = time of flight (can be calculated using the horizontal motion)

The horizontal motion is not affected by gravity and can be described by the equation:

x = vi_x * t

where
vi_x = initial horizontal velocity (12 m/s, given)

We know that the horizontal distance the dart travels is 2.7 m, so we can use the equation above to calculate the time of flight:

2.7 m = 12 m/s * t
t = 2.7/12 s = 0.225 s

Then we can substitute this value of t into the equation for vertical motion to calculate the distance the dart falls below the intended target:

y = 0 + 1/2 * 9.8 m/s^2 * (0.225 s)^2
y = 0.038 m

So the dart falls 0.038 m below the intended target.

Answer 2

To solve this problem, we can use the equations of motion for projectile motion. Since the dart is thrown horizontally, we know that there is no vertical velocity and that the only acceleration acting on the dart is gravity, which is 9.8 m/s^2 downward.

The vertical motion of the dart can be described by the equation:

y = y0 + v0t + 0.5at^2

where y is the vertical position of the dart, y0 is the initial vertical position (which is 0 in this case), v0 is the initial vertical velocity (which is also 0), t is the time of flight, and a is the acceleration due to gravity.

To find the time of flight, we can use the horizontal motion of the dart, which is described by the equation:

x = x0 + v0t + 0.5at^2

where x is the horizontal position of the dart, x0 is the initial horizontal position (which is 0), v0 is the initial horizontal velocity (which is 12 m/s), and t is the time of flight.

We know that the horizontal position of the dart is 2.7 m, so we can substitute that into the equation and solve for t:

2.7 = 0 + 12t

t = 2.7/12 = 0.225 seconds

We can then substitute t and a into the vertical motion equation to find the vertical position of the dart:

y = 0 + 0 + 0.5(9.8)(0.225^2)

y = -0.061 m

So the dart hits 0.061 m below the intended target.