A rescue plane wants to drop supplies to isolated mountain...

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge H = 237 m below. If the plane is traveling horizontally with a speed of 239 km/hr (66.4 m/s), how far in advance of the recipients (horizontal distance) must the goods be dropped? Suppose, instead, that the plane releases the supplies a horizontal distance of x = 403 m in advance of the mountain climbers. What vertical velocity (use the positive direction as upwards) should the supplies be given so that they arrive precisely at the climbers' position? With what speed do the supplies land in the latter case?

Transcript

00:01 Hi, here in this given problem this is the height of the plane above the climber, a solo climber.

00:16 This is the direction, horizontal direction of the motion of the aeroplane and the packet released by the aeroplane.

00:27 It is received by the climber here at this point at a distance x from the aeroplane.

00:39 The height of the plane above the climber that is given as 237 meter.

00:48 Its horizontal velocity v0, that is 66 .4 meter per second.

00:56 So first of all, we find time taken by the packer.

01:00 To go up to the ground for which we will be using second equation of motion h is equal to v -o -y initial vertical velocity of the package into t plus half g t square but v -o -y here that is zero in this case for h 237 is equal to 0 plus half into 9 .8 into t square time we have to find so, that time is given by square root of 2 times of 237 divided by 9 .8, and it is calculated to be equal to 6 .95 second.

01:44 So the horizontal distance x, as the horizontal motion remains uniform, so distance equals to speed into time, means that is 6 .6 ,000 multiplied by the time which is 6 .95, and it is calculated to be equal to 461 .8 meter, which is the answer, one of the answer, first answer for this given problem here means 461 .8 meter is the distance behind the recipient.

02:23 Now, in the second part of the problem, this time the distance, it becomes just 400.

02:35 103 meter means it becomes less.

02:38 So we should drop the package a little vertical also, means in a slant direction.

02:46 So first of all we find time covered, time taken by the package to go up to this much distance with the same horizontal velocity.

02:54 So that will be given by 403.

02:57 Time equals to distance upon speed.

02:59 Speed vox that will be taken as same as that of the plane.

03:04 Means 66 .4.

03:07 So this time comes out to be equal to 6 .07 second...