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For the ballistic missile aimed to achieve the maximum range of 9500 km, what is the maximum altitude reached in the trajectory

User Zeokav
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2 Answers

28 votes
28 votes

Final answer:

Without specific details such as initial velocity or launch angle, it is not possible to calculate the precise maximum altitude reached by a missile with a range of 9500 km. Real-world ballistic missile trajectories are complex and often reach suborbital altitudes, far surpassing conventional projectiles.

Step-by-step explanation:

The question regarding the ballistic missile achieving a maximum range of 9500 km and its maximum altitude falls under the subject of Physics, specifically projectile motion and kinematics. In an idealized scenario without air resistance, the maximum altitude (or maximum height) reached by a projectile depends on the initial speed and the angle of launch. However, the question does not provide enough information such as the launch angle, initial velocity, or whether the missile's engines provide constant thrust or not. Thus, calculating the precise maximum altitude reached by a missile achieving the range of 9500 km is not possible without additional information. For general understanding, if we considered the missile to be a simple projectile (which is not the case for intercontinental ballistic missiles), its maximum altitude could be found using kinematic equations, assuming the optimum angle of 45 degrees for maximum range in vacuum.

Various factors in real-world scenarios complicate this calculation, including the effect of the Earth's atmosphere, the curvature of the Earth, and the propulsion dynamics of the missile. Intercontinental ballistic missiles (ICBMs) also typically achieve suborbital spaceflight, reaching altitudes much higher than conventional projectiles, sometimes exceding the boundary of space at 100 km above Earth's surface (the Kármán line) before descending towards the target.

User Ngearing
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12 votes
12 votes

Step-by-step explanation:

The range R of a projectile is given the equation


R = (v_0^2)/(g)sin(2\theta)

The maximum range is achieved when
\theta = 45° so our equation reduces to


R_(max) = (v_0^2)/(g)

We can solve for the initial velocity
v_0 as follows:


v_0^2 = gR_(max) \Rightarrow v_0 = \sqrt{gR_(max)}

or


v_0 = \sqrt{(9.8\:\text{m/s}^2)(9.5×10^6\:\text{m})}


\:\:\:\:\:\:\:=9.6×10^3\:\text{m/s}

To find the maximum altitude H reached by the missile, we can use the equation


v_y^2 = v_(0y)^2 - 2gy = (v_0sin(45°))^2 - 2gy

At its maximum height H,
v_y = 0 so we can write


0 = (v_0sin(45°))^2 - 2gH

or


H = ((v_0sin(45°))^2)/(2g)


\:\:\:\:\:\:= \frac{[(9.6×10^3\:\text{m/s})sin(45°)]^2}{2(9.8\:\text{m/s}^2)}


\:\:\:\:\:\:= 2.4×10^6\:\text{m}

User Feetwet
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