General Physics, 2 questions

A relief plane is delivering a food package to a group of people stranded on a very small island. the island is too small for the plane to land on and the only way to deliver the package is by dropping it. The airplane flies horizontally with constant speed $v0$ and altitude $h0$. The positive x and y directions are defined in the figure.

part A) Find the initial velocity of the package with respect to the ground.

part B) How long it will take the package to reach sea level from the time it is ejected?

part C) If the package were to land right on the island, at what horizontal distance $D$ from the plane to the island should the package be released.

part D) What is the horizontal distance $D_h$ from the plane to the island when the package hits the ground.

part E) Find the velocity $v_f$ of the package when it hits the ground.

part F) The damage to the package decreases with decreased impact speed. What ejection speed $v$ minimizes the damage done to the package?

General Physics Questions



the initial speed relative to the ground is $U0 = V0-V1$

toward the positive direction of x


there is no initial speed on the y axis. On the y axis there is free fall.

(h is the initial height, g is the gravitational acceleration)

$h = g*t^2/2$

$t = sqrt{(2*h/g)}$


on the x axis there is no acceleration. It implies

$D = U0*t = (V0-V1)*sqrt{(2*h/g)}$


The distance traveled by the plane in time t is

$d = V0*t = V0*sqrt{(2*h/g)}$

The distance from island is

$D1 = D-d = v1*sqrt{(2*h/g)}$


On the x axis $V rx = U0$, on the y axis $V ry = V y$

where $Vy^2 =2*g*h$,   $Vy=sqrt{2gh}$
$Wr = sqrt{Vrx^2+Vry^2} = sqrt{U0^2 + 2gh} = sqrt{V0-V1)^2 + 2gh}$

$tan(alpha) = Vry/vrx = (2gh)/(V0-V1)$

where $alpha$ is the angle between $Wr$ and $x$ axis.


the condition for minimum damage is $Wr$ is minimum.

$dWr/dV1 =0$

$dW1/dV1 = -2*(V0-V1)/[2(sqrt{(V0-V1)^2 +2gh)}]$

This happens when $V1= V0$ and the value of $Wr$ is $Wr = sqrt{2*g*h}$

General Physics Second Question

$r = [3.6 cm +3 (cm/s/s) *t^2]*i + [5.5 (cm/s)*t]*j$

a) $r(0) = 3.6*i cm$

$r(2) =(3.6 +3*4)*i +5.5*2*j = (15.6*i +11*j) cm$

$r(2)-r(0) = (12*i +11*j) cm$

average speed is

$V = (r(2)-r(0)/(2-0) =(6*i +5.5*j) cm/s$

$|V| = sqrt{6^2 +5.5^2} = 8.14 cm/s$

$tan(alpha) = Vy/Vx =5.5/6 = 0.916$

$alpha =42.51$ deg angle from the x axis to the vector v


$v = dr/dt = 6*t*i +5.5*j cm/s$

$v(0)= 5.5*j cm/s$

$v(0) = (0, 5.5) cm/s$


$v(1) = 6*i +5.5*j  cm/s$

$v(1) = (6, 5.5) cm/s$


$v(2)= 6*2*i +5.5*j = 12*i +5.5*j cm/s$

$v(2)= (12, 5.5) cm/s$