A dart is thrown at a dartboard 2.37 m away. When the dart is released at the same height as the center of the dartboard, it hits the center 0.447 sec. At what angel relative to the floor was the dart thrown.
Thanks :)
To find the angle at which the dart was thrown, we can use the horizontal and vertical components of its motion.
Step 1: Determine the horizontal velocity of the dart.
We know that the dart hits the center of the dartboard at a distance of 2.37 m in 0.447 seconds. The dart travels horizontally, so we can determine its horizontal velocity using the formula:
horizontal velocity = distance / time
horizontal velocity = 2.37 m / 0.447 s = 5.30 m/s
Step 2: Determine the vertical velocity of the dart.
Since the dart is released at the same height as the center of the dartboard, the initial vertical velocity is zero. Therefore, the only vertical velocity component of the dart is due to the effect of gravity. We can find the vertical velocity using the formula:
vertical velocity = (change in vertical position) / time
vertical velocity = 0 m/s - (-9.8 m/s^2) * 0.447 s
vertical velocity = 4.38 m/s
Step 3: Calculate the angle.
We can use the tangent function to determine the angle:
tan(angle) = vertical velocity / horizontal velocity
angle = arctan(vertical velocity / horizontal velocity)
angle = arctan(4.38 m/s / 5.30 m/s)
angle ≈ 40.06 degrees
Therefore, the dart was thrown at an angle of approximately 40.06 degrees relative to the floor.
To determine the angle at which the dart was thrown, we can use the equations of horizontal and vertical motion.
First, let's find the initial velocity of the dart. We can use the equation of vertical motion:
h = ut + 0.5 * gt^2
Where h is the height, u is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight. Since the dart was released at the same height as the center of the dartboard, h = 0.
0 = (u * 0.447) + 0.5 * 9.8 * (0.447)^2
Solving this equation will give us the initial vertical velocity of the dart.
Next, we need to find the horizontal velocity of the dart. We can use the equation of horizontal motion:
s = ut
Where s is the horizontal distance, u is the initial horizontal velocity, and t is the time of flight.
Rearranging the equation to solve for u:
u = s / t
Substituting the given values:
u = 2.37 / 0.447
This will give us the initial horizontal velocity of the dart.
Now, we can find the angle using the relationship between the horizontal and vertical components of velocity.
tan(theta) = u_vertical / u_horizontal
Substituting the previously calculated values:
tan(theta) = (Initial vertical velocity) / (Initial horizontal velocity)
Using the inverse tangent function, we can find the angle:
theta = atan((Initial vertical velocity) / (Initial horizontal velocity))
Calculating this expression will give us the angle at which the dart was thrown relative to the floor.
t = sqrt(2h/9.8)
0.447s = 0.452 H
H = 0.988
now find the angle,
use your calculator and plug in
arctan (0.988 / 2.37 )
should be 22.6 degrees.