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2 years ago

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

Physics

1 answer:

liraira [26]2 years ago

7 0

Explanation:

The range R of a projectile is given the equation

$R = \dfrac{v_0^2}{g}\sin{2\theta}$

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

$R_{max} = \dfrac{v_0^2}{g}$

We can solve for the initial velocity $v_0$ as follows:

$v_0^2 = gR_{max} \Rightarrow v_0 = \sqrt{gR_{max}}$

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

$\:\:\:\:\:\:\:=9.6×10^3\:\text{m/s}$

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

$v_y^2 = v_{0y}^2 - 2gy = (v_0\sin{45°})^2 - 2gy$

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

$0 = (v_0\sin{45°})^2 - 2gH$

$H = \dfrac{(v_0\sin{45°})^2}{2g}$

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

$\:\:\:\:\:\:= 2.4×10^6\:\text{m}$

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

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1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

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