A dart is thrown horizontally with an initial speed of 11 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?
(a) Well, PQ is quite the mystery. It's like the dart wanted a scenic route. But fear not, my dear friend! To find PQ, we need to know how far the dart traveled horizontally in 0.19 seconds. Luckily, we know the initial horizontal velocity is 11 m/s, so we can use the formula distance = velocity × time. That means PQ = 11 m/s × 0.19 s. Crunch those numbers and you'll have the answer!
(b) Ah, the distance from the dart board... It's the ultimate getaway spot for darts. To find this distance, we need to figure out how long the dart was in the air. Luckily, we already know it took 0.19 seconds to reach point Q. So, we can use the equation distance = velocity × time to find the distance. The only catch is that the velocity we're looking for is the initial horizontal velocity, which is given as 11 m/s. So, let's plug in the numbers and see how far away our adventurous dart was released!
To solve this problem, we need to use the principles of projectile motion. Let's break down the problem step by step.
(a) To find the distance PQ, we first need to determine the vertical component of the dart's velocity when it hits point Q.
We know that the horizontal component of the velocity is constant throughout the motion because the dart is thrown horizontally. So, the horizontal component of velocity is 11 m/s.
Now, we need to find the vertical component of velocity. We can use the equation:
vf = vi + at
Where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial velocity (vi) in the vertical direction is zero because the dart is not thrown vertically.
The acceleration (a) is due to gravity and is approximately 9.8 m/s² downward.
The final velocity in the vertical direction (vf) can be found by rearranging the equation:
vf = vi + at
vf = 0 + (9.8 m/s²)(0.19 s)
vf = 1.862 m/s
So, the vertical component of the velocity when it hits point Q is approximately 1.862 m/s downward.
Now, we can find the distance PQ using the formula:
PQ = vf × t
PQ = (1.862 m/s)(0.19 s)
PQ = 0.353 m (rounded to three decimal places)
Therefore, the distance PQ is approximately 0.353 meters.
(b) To find how far away from the dart board the dart is released, we can use the formula for horizontal distance traveled during projectile motion:
d = vi × t
where d is the horizontal distance, vi is the initial horizontal velocity, and t is the time.
In this case, the initial horizontal velocity (vi) is 11 m/s, and the time (t) is given as 0.19 s.
So, the distance away from the dart board when the dart is released is:
d = (11 m/s)(0.19 s)
d = 2.09 meters (rounded to two decimal places)
Therefore, the dart is released approximately 2.09 meters away from the dart board.