A dart is thrown horizontally with an initial speed of 14 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.19 s later. (a) What is the distance PQ? (b) How far away from the dart board is the dart released?
how far does it fall in .19 s ?
delta h = (1/2) g t^2
= 4.9*.19^2 = .177 meter = 17.1 cm
u = 14
d = u t = 14 *.19 = 2.66 m to board
To solve this problem, we can use the equations of motion in the x and y directions separately.
(a) What is the distance PQ?
We know that the vertical distance (PQ) covered by the dart in 0.19 s is the result of the downward acceleration due to gravity. We can use the following kinematic equation to find the vertical distance:
∆y = v_y * ∆t + (1/2) * g * ∆t^2
Here,
∆y = PQ (vertical distance covered)
v_y = 0 (initial vertical velocity since the dart was thrown horizontally)
∆t = 0.19 s (time taken)
g = 9.8 m/s^2 (acceleration due to gravity)
∆y = 0 * 0.19 + (1/2) * 9.8 * (0.19)^2
∆y = 0.09014 m (rounded to four decimal places)
Therefore, the distance PQ is approximately 0.0901 meters.
(b) How far away from the dart board is the dart released?
Since the dart is thrown horizontally, there is no acceleration in the horizontal direction. Therefore, the horizontal distance traveled by the dart is equal to the product of the horizontal velocity (which remains constant) and the time taken, as given by:
∆x = v_x * ∆t
Here,
∆x = distance traveled in the horizontal direction
v_x = initial horizontal velocity = 14 m/s
∆t = 0.19 s
∆x = 14 * 0.19
∆x = 2.66 m (rounded to two decimal places)
Therefore, the dart is released approximately 2.66 meters away from the dart board.
To solve this problem, we can use basic principles of projectile motion and kinematics. Let's break it down step by step:
(a) What is the distance PQ?
1. First, we need to find the horizontal distance covered by the dart in 0.19 seconds.
- The horizontal component of velocity remains constant throughout the motion since there is no horizontal acceleration. Therefore, the horizontal speed of the dart is 14 m/s.
- We can use the formula: distance = speed × time. Thus, the horizontal distance covered by the dart is: distance = 14 m/s × 0.19 s.
2. Now, we have to find the vertical distance traveled by the dart during this time.
- Since the dart is thrown horizontally, there is no initial vertical velocity. The only acceleration acting on the dart is due to gravity, which accelerates the dart downwards.
- We can use the formula: vertical distance = (1/2) × acceleration × time².
- The acceleration due to gravity is approximately 9.8 m/s². Therefore, the vertical distance traveled by the dart is: vertical distance = (1/2) × 9.8 m/s² × (0.19 s)².
3. Finally, to find the distance PQ, we can use the Pythagorean theorem since PQ is the hypotenuse of a right triangle formed by the horizontal and vertical distances.
- The Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
- Mathematically, it is expressed as: (distance PQ)² = (horizontal distance)² + (vertical distance)².
- Plug in the values we found earlier and solve for (distance PQ).
(b) How far away from the dart board is the dart released?
To determine the initial horizontal distance from the dart board to the release point, we consider the time it takes for the dart to hit the target.
1. We know that the horizontal speed of the dart remains constant throughout its motion and is equal to 14 m/s.
2. The formula for horizontal distance is: distance = speed × time.
- Rearrange the formula to solve for time: time = distance / speed.
- In this case, distance is the distance between the release point and the target, and speed is the horizontal speed of the dart (which is 14 m/s).
- Substitute the values and calculate the time.
3. The time calculated in step 2 represents the time taken by the dart to hit the target.
4. By multiplying this time by the horizontal speed of 14 m/s, we can find the horizontal distance the dart traveled before hitting the target, which is the initial distance from the dart board to the dart release point.
Keep in mind that the values used for acceleration due to gravity (9.8 m/s²) and the initial horizontal speed (14 m/s) are approximate and can slightly vary depending on the environment and conditions.