A dart is thrown horizontally with an initial speed of 6 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.29 s later. Neglect air resistance.
(a) What is the distance PQ?
(b) How far away from the dart board is the dart released?
The answer to (b) is 1.74 m. Please help me in determining what (a) is.
how far does a dart fall in free fall in .29sec?
h=1/2 g t^2
thank you!
To determine the distance PQ, we need to use the kinematic equations of motion. Let's break down the problem step by step:
Step 1: Determine the time of flight
Since the dart is thrown horizontally, it only travels vertically due to gravity. The time of flight can be calculated using the formula:
time = (2 * vertical distance) / gravity
Given that the time is 0.29 s and neglecting air resistance (which means the only force acting on the dart is gravity), we can rewrite the formula as:
0.29 s = (2 * vertical distance) / 9.8 m/s²
Solving for the vertical distance:
vertical distance = (0.29 s * 9.8 m/s²) / 2
vertical distance = 1.421 m
Step 2: Determine the horizontal distance
Since the dart was thrown horizontally, its initial horizontal velocity remains constant throughout its motion. We can use the formula for horizontal distance traveled:
horizontal distance = horizontal velocity * time
Given that the initial horizontal velocity is 6 m/s and the time is 0.29 s, we can calculate the horizontal distance:
horizontal distance = 6 m/s * 0.29 s
horizontal distance = 1.74 m
Step 3: Determine the distance PQ
To determine the distance PQ, we can use the Pythagorean theorem, as PQ is the hypotenuse of a right triangle with horizontal and vertical distances as the other two sides:
PQ = sqrt((horizontal distance)² + (vertical distance)²)
Substituting the values we have:
PQ = sqrt((1.74 m)² + (1.421 m)²)
PQ = sqrt(3.0276 m² + 2.018841 m²)
PQ = sqrt(5.046441 m²)
PQ = 2.24 m
Therefore, the distance PQ is approximately 2.24 meters.