An airplane with a speed of 71.2m/s is climbing upward at an angle of 60.8 degrees with respect...

Question:

An airplane with a speed of 71.2m/s is climbing upward at an angle of 60.8 degrees with respect to the horizontal. When the plane's altitude is 711m, the pilot releases a package.

a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

Projectile Motion:

Projectile motion is the parabolic motion of a body on which the only acceleration is gravity. The acceleration in the horizontal motion of a projectile is zero. Because the motion characteristics are different, we calculate the displacement, velocity, and angle of displacement or velocity at any point by first calculating them separately and then taking the vector equivalent.

Question (a)

This question asks us to calculate the range of the projectile. This means the displacement in the horizontal axis, for which we know the horizontal speed but do not know the time of flight. We can only determine the time taken for this by considering the vertical motion as follows:

From the date given:

• The vertical displacement is {eq}- 711m{/eq} because it's in the direction opposite the initial velocity.
• Initial velocity {eq}u_y = 71.2\sin 60.8 = 62.2 \ m/s{/eq}

Choosing an appropriate equation from Newton's Laws, we get the time t as:

{eq}\begin{align*} s &= ut - \frac{1}{2}gt^2\\ -711 &= 62.2t - 0.5(9.81)t^2\\ 4.905t^2 - 62.2t - 711&= 0\\ t &= \frac{62.2 \pm \sqrt{62.2^2 - 4(4.905)(-711)}}{2(4.905)}\\ &= 19.9s \end{align*}{/eq}

The distance required s is the product of horizontal velocity and time:

{eq}s = v_xt\\ s = 71.2 \cos 60.8 \times 19.9 \\ \boxed{s = 702 \ m}{/eq}

The package will hit the Earth 702 meters away from the point directly below the point of release.

Question (b)

To determine the angle of the velocity before impact, let's determine the components of the final velocity first. The horizontal component does not change, so we need not worry about that. Because we now have the time of flight t, let's use that to determine by choosing an appropriate equation from Newton's laws the final velocity {eq}v_y {/eq} in the vertical axis:

{eq}\begin{align*} v_y &= u_y - gt\\ &= 62.2 - 9.81(19.9)\\ &= 133.2 \ m/s \end{align*}{/eq}

We determine the required angle by taking the {eq}\tan\theta{/eq} ratio of the components as follows:

{eq}\begin{align*} \tan \theta &= \frac{v_y}{u_y}\\ \theta&= \arctan\frac{133.2}{71.2 \cos 60.8 }\\ \theta&= \boxed{75.4^\circ} \end{align*}{/eq}

Therefore, the angle, taken relative to the ground, of the final velocity is 75.4 degrees. 