Answer
$(a)\space 239\space m/s,\space \alpha=57.1^{\circ}$ with respect to the horizontal
$(b)\space 239\space m/s$ at an angle of $57.1^{\circ}$ with respect to the horizontal.
Work Step by Step
(a) Here we need to find the x,y-velocity components of the fuel tank just before it hits the ground. x-component of the velocity remains constant (x-component of the velocity of the plane). Because there isn't acceleration in the horizontal direction, we can write,
$V_{x}=Vcos\theta=135\space m/s\times cos15^{\circ}$
$V_{x}=130\space m/s$
Let's apply the equation $V^{2}=u^{2}+2aS$ in the vertical direction to find the y-component of the velocity of the plane.
$\uparrow V_{y}^{2}=u^{2}+2aS$ ; Let's plug known values into this equation.
$V_{y}^{2}=(135\space m/s\times sin15^{\circ})^{2}+2(-9.8\space m/s)(-2000\space m)$
$V_{y}^{2}=1221\space m^{2}/s^{2}+39200\space m^{2}/s^{2}=40421\space m^{2}/s^{2}$
$V_{y}=-201\space m/s$
By using the Pythagorean theorem, we can get.
Magnitude of the velocity of the fuel tank $V=\sqrt {V_{x}^{2}+V_{y}^{2}}$
$V=\sqrt {(130\space m/s)^{2}+(201\space m/s)^{2}}=239\space m/s$
By using trigonometry, We can get the inclined angle $\alpha$ of the velocity vector with respect to the horizontal.
$tan\alpha=\frac{V_{y}}{V_{x}}=\frac{201\space m/s}{130\space m/s}$
$\alpha=tan^{-1}(1.55)=57.1^{\circ}$
(b) The velocity of the fuel tank in part (b) just before impact is 239 m/s at an angle $57.1^{\circ}$ with respect to the horizontal.
