A gun is accurately aimed at a dangerous criminal hanging from the gutter of a building. The target is well within the gun’s range, but the instant the gun is fired and the bullet moves with a speed v(sub0) , the criminal lets go and drops to the ground. What happens? The bullet:
1) hits the criminal regardless of the value v(sub0)
2) hits the criminal only if v(sub0) is large enough
3) misses the criminal.
If you were to draw the triangle, width would be 3m, the height would be 2m, and the hypotenuse would be 2^2+3^2=13?
I don't know how to prove that the bullet will hit the criminal with equations. please help! thank you!
THIS IS OLD... Y'AAL MUST BE OLD BY NOW
1)
the bullet accelerates down just as fast as the criminal accelerates down.
yes, but how would i prove that with equations?
Both are in the air for time t
How far does bullet drop from straight line in time t ?
The only vertical force on the bullet is m g
so it's height as a function of time is
h = (Vo sin T) t - (1/2) g t^2
but Vo sin T t is the straight line to the target ignoring gravity
so it drops (1/2) g t^2
so does the criminal
both of them move v=gt at the same rate.
thank you!
To determine what happens to the bullet when the criminal lets go, we need to analyze the motion of both the bullet and the criminal separately. Let's break it down step by step:
1) First, let's consider the motion of the criminal. When they let go of the gutter, they start falling freely under the influence of gravity. The time it takes for the criminal to hit the ground can be determined using the equation for free fall:
h = (1/2) * g * t^2
Where:
h = height of the building (2m in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time of fall
Rearranging the equation, we get:
t = sqrt((2h) / g)
Plugging in the values, we find:
t = sqrt((2 * 2) / 9.8) = sqrt(4 / 9.8) = sqrt(0.408) ≈ 0.64 seconds
So, it takes approximately 0.64 seconds for the criminal to hit the ground.
2) Now let's analyze the motion of the bullet. Since the gun is accurately aimed at the criminal, we assume that the bullet is initially fired in the same direction as the target. However, once the bullet leaves the gun, it has its own velocity and is no longer influenced by the gun.
To determine whether the bullet hits the criminal, we need to compare the time it takes for the bullet to reach the criminal with the time it takes for the criminal to fall to the ground.
The distance the bullet travels horizontally is equivalent to the width of the triangle (3m in this case). The time it takes for the bullet to reach the criminal can be found using the equation:
t_bullet = d / v_0
Where:
d = distance traveled by the bullet horizontally (3m in this case)
v_0 = initial velocity of the bullet (speed at which it was fired)
To determine whether the bullet hits the criminal, we need to compare t_bullet with the time it takes for the criminal to fall (approximately 0.64 seconds).
If t_bullet is less than or equal to the time taken by the criminal to fall (0.64 seconds), then the bullet will hit the criminal. Otherwise, it will miss.
So, in conclusion, to determine the fate of the criminal, we compare the time it takes for the bullet to reach the criminal (t_bullet) with the time it takes for the criminal to fall (approximately 0.64 seconds). If t_bullet ≤ 0.64 seconds, the bullet hits the criminal, and if t_bullet > 0.64 seconds, the bullet misses the criminal.
Keep in mind that this analysis assumes no air resistance, no other forces acting on the bullet or the criminal, and that the bullet is initially fired with an accurate aim.