A rifle is aimed horizontally at a target 30 m away. The bullet hits the target 1.9 cm below the aiming point. What are the a) the bullet's time of flight and b) its speed as it emerges from the rifle?
A) is expose to be 62ms
B) is expose to be 480m/s
this is from by book. But I am getting wrong answers What am I doing wrong?
How long does it take to fall 1.9 cm?
d=1/2 g t^2 solve for time.
Now, that is the same time it took to go 30 m horizontally. Figure speed.
To calculate the bullet's time of flight, we can start by determining the time it takes for the bullet to fall 1.9 cm. We can use the equation d = (1/2)gt^2, where d is the distance (1.9 cm in this case), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Since we want to find time, we rearrange the equation to solve for t:
t^2 = 2d/g
t = √(2d/g)
Substituting the values, we have:
t = √[(2 * 0.019 m) / 9.8 m/s^2] = √(0.038/9.8) = √0.0039 ≈ 0.062 s (rounded to 3 decimal places)
So, the bullet's time of flight is approximately 0.062 seconds.
Now, let's calculate the bullet's speed as it emerges from the rifle. Since the bullet travels horizontally for a distance of 30 m in the same time it takes to fall 1.9 cm vertically, we can determine its speed by dividing the horizontal distance by the time:
Speed = Distance / Time = 30 m / 0.062 s = 484.31 m/s (rounded to 3 decimal places)
Therefore, the correct answers are:
a) The bullet's time of flight is approximately 0.062 seconds (62 ms).
b) The bullet's speed as it emerges from the rifle is approximately 484.31 m/s.
If your answers are different, please check your calculations and ensure that you input the correct values for the variables in the equations.