A fighter-jet is in level flight at an altitude of 400m and with a speed of 200m/s. The fighter launches a rocket-powered missile horizontally from the underside of the jet. The rocket engine provided a constant horizontal acceleration of 25.0 m/s^2.
a) how far does the missile travel horizontally from the launch point to impact with the ground?
b) calculate the components of the missile's impact velocity.
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To answer these questions, we can use the equations of motion for linear motion. We'll need to consider the horizontal and vertical motion of the missile separately.
a) To find the horizontal distance the missile travels, we need to calculate the time it takes for the missile to hit the ground. We can use the equation:
y = ut + (1/2)at^2
where:
y = vertical displacement (altitude) = -400m (negative because it is in the downward direction)
u = initial vertical velocity = 0 (since the missile is launched horizontally)
a = acceleration in the vertical direction = acceleration due to gravity = -9.8 m/s^2 (negative because it is in the downward direction)
t = time
Rearranging the equation, we get:
t^2 - (2u/a)t - (2y/a) = 0
Substituting the given values, we have:
t^2 - (2(0)/-9.8)t - (2(-400)/-9.8) = 0
t^2 + 81.63 = 0
Solving the quadratic equation, we find two solutions for t: t = √(81.63) = 9.04s and t = -√(81.63) = -9.04s. We discard the negative value since time cannot be negative.
Therefore, it takes approximately 9.04 seconds for the missile to hit the ground.
Now, we can calculate the horizontal distance traveled using the equation:
s = ut + (1/2)at^2
In this case, the initial horizontal velocity (u) of the missile is the same as the fighter jet's velocity, which is 200m/s. Additionally, the horizontal acceleration (a) is given as 25.0m/s^2.
Substituting the values, we have:
s = (200)(9.04) + (1/2)(25)(9.04)^2
s = 1808 + (0.5)(25)(81.632)
s = 1808 + 10204
s = 12012 m
Therefore, the missile travels approximately 12012 meters horizontally from the launch point to impact with the ground.
b) To calculate the components of the missile's impact velocity, we can use the equations of motion for linear motion.
The horizontal component of the impact velocity will remain constant throughout the motion. So, the horizontal component of the impact velocity (Vx) is the initial horizontal velocity (200m/s).
For the vertical component, we can use the equation:
v = u + at
where:
v = final vertical velocity
u = initial vertical velocity = 0 (since the missile is launched horizontally)
a = acceleration in the vertical direction due to gravity = -9.8 m/s^2
Substituting the values, we have:
v = 0 + (-9.8)(9.04)
v = -88.592 m/s
Therefore, the components of the missile's impact velocity are:
Vx = 200 m/s (horizontal component)
Vy = -88.592 m/s (vertical component, downward direction)