Question

Relative Velocity and Projectile Motion . a)An airplane moves horizontally with a constant velocity VA/G 300m/s with respect to ground at height H = 600m. The plane fires a container, which has a speed VC/A = 300m/s and direction α = 300 with respect to the airplane immediately after the firing. The container performs a projectile motion and hits the ground at a horizontal distance L from the point of the firing (see figure a)What is the velocity (horizontal and vertical components) of the container immediately after the firing with respect to the ground Vc/G (10 pt) b) Find the distance L at which the container hits the ground. (14 pt) c) Does the time of the flight depend on the angle a? Explain. 2 pt d) what is acceleration of the container at the maximum height of it's trajectory with respect to the ground (2 points) e) what is acceleration of the container at the maximum height of it's trajectory with respect to the airplane, which is moving with constant velocity. (2 points)

Answer 1

Positive X axis is horizontally along the initial direction of motion of airplane. Positive Y axis is vertically upwards.

Velocity of airplane with respect to ground .

Positive X axis is horizontally along the initial direction of motion of airplane. Positive Y axis is vertically upwards. Velocity of airplane with respect to ground V_A/G = 300 x^cap m/s Velocity of container with respect to airplane V_C?A = 300 (cos 30 degree x^cap + sin 30 degree y6cap) m/s The container during it's journey through air, goes through a horizontal displacement of L x^cap and a vertical displacement of_h y^cap = -600 y^cap m (a) Using relative velocity formula V_C/G = V_C/A + V_A/G V_C/G = V_C/A + V_A/G = 300 (cos 30 degree x^cap + sin 30 degree y^cap) + 300 x^cap = [300 + 300 cos 30 degree) x^cap + 300 sin 30 degree y^cap]m/s = [559.81 x^cap + 150 y^cap] m/s Hence immediately after firing, horizontal and vertical components are 559.81 m/s and 150 m/s respectively (b) Using the equation delta r = V_0 t + 1/2 at^2 delta y = v_0yt + 1/2 a_y t^2 and delta x = V_0 x t + 1/2 a_x t^2

Velocity of container with respect to airplane $\mathbf{V}_{C/A}=300(\cos30\degree\hat{x}+sin30\degree\hat{y})\,m/s$

The container during it's journey through air, goes through a horizontal displacement of $L\hat{x}$ and a vertical displacement of $-H\hat{y}=-600\hat{y}\,m$ .

(a)

Using relative velocity formula $\mathbf{V}_{C/G}=\mathbf{V}_{C/A}+\mathbf{V}_{A/G}$

$\mathbf{V}_{C/G}=\mathbf{V}_{C/A}+\mathbf{V}_{A/G}$

$=300(\cos30\degree\hat{x}+sin30\degree\hat{y})+300\,\hat{x}=\left[(300+300\cos30\degree)\hat{x}+300\sin30\degree\,\hat{y} \right ]\,m/s$

$=\left[559.81\hat{x}+150\hat{y} \right ]\,m/s$

Hence immediately after firing, horizontal and vertical components are and respectively.

(b)

Using the equation $\Delta{\mathbf{r}}=\mathbf{V}_0t+\frac{1}{2}\mathbf{a}t^2$

$\Delta{y}=V_{0y}t+\frac{1}{2}a_yt^2$ and $\Delta{x}=V_{0x}t+\frac{1}{2}a_xt^2$

$-600=150t+\frac{1}{2}(-9.81)t^2$

$9.81t^2-300t-1200=0$

Solving the above equation, $t=34.162\,s$

During the same time, $L=(559.81)t+\frac{1}{2}(0)t^2=(559.81)(34.162)=19124.23\,m$

(c)

To find the time of flight, we used the equation $-H=V_{0}\sin\alpha t+\frac{1}{2}(-g)t^2$

$gt^2-2V_{0}\sin\alpha t-2H=0$

Time of flight $t=\frac{2V_0\sin\alpha\pm\sqrt{4V^2_0\sin^2\alpha+8gH}}{2g}$

$t=\frac{V_0\sin\alpha+\sqrt{V^2_0\sin^2\alpha+2gH}}{g}$

Hence time of flight depends upon angle $\alpha$

(d)

Acceleration of container is constant. It is always the same at any point of it's trajectory. $\mathbf{a}_{C/G}=-g\hat{y}$ .

Hence acceleration of container at the maximum height of the trajectory with respect to ground is $|\mathbf{a}_{C/G}|=g=9.81\,m/s^2$ vertically downwards.

(e)

Using the relative acceleration formula, $\mathbf{a}_{C/G}=\mathbf{a}_{C/A}+\mathbf{a}_{A/G}$

Since airplane is moving with constant velocity, $\mathbf{a}_{A/G}=0$ .

$\mathbf{a}_{C/A}=\mathbf{a}_{C/G}=-g\hat{y}$

This acceleration also is constant.

Hence at the maximum height of the trajectory, acceleration of container with respect to the airplane is $|\mathbf{a}_{C/A}|=g=9.81\,m/s^2$ vertically downwards.