What is the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean? answer in units of m/s.A plane drops a hamper of medical supplies from a height of...

What is the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean? answer in units of m/s.

A plane drops a hamper of medical supplies from a height of 2690 m during a practice run over the ocean. The plane's horizontal velocity was 131 m/s at the instant the hamper was dropped. the acceleration of gravity is 9.8 m/s^2.

Asked on by jayjay00

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krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

The approach of the answer is correct, but it fails to recognize the difference between speed and velocity. AS a result the answer is incomplete. The speed measures the magnitude of rate of change of position. It does not consider the direction of movement. However, velocity specifies in addition to the magnitude of rate of change of position, which is same as speed, also the direction of movement. The answer given above does not specify the direction of movement.

I am giving below a complete solution.

In solving this problem we assume that at the time of dropping the hamper was flying horizontally. Therefore at the time of dropping the hamper its velocity was equal to the velocity of the plane. This is 131 m/s in horizontal direction. We further assume that throughout its fall the hamper does not experience any air resistance therefore the horizontal component of its velocity continues to be 131 m/s.

Just as the hamper is dropped from plane, it also begins to develop vertical velocity under the influence of gravity. The rate of this acceleration in vertical velocity is given as 9.8 m/s^2.

As the hamper leaves the plane it has net horizontal velocity equal to 131 m/s. But as it falls not only the vertical component of velocity develops, the direction of of net velocity also changes. with slowly increasing angle from the initial direction.

We will fist calculate the magnitude of the velocity of the hamper as it hits the water surface. Then we will determine the angle which net velocity makes with the horizontal.

Initially we will calculate the vertical component of velocity.

For an accelerating body the relationship between its velocity, acceleration and distance travelled is given by:

v^2 = u^2 + 2as.

where:

v = final velocity

u = initial velocity = 0

a = acceleration = 9.8 m/s^2

s = distance travelled = 2690 m

Therefore: v^2 = 0 + 2*9.8*2690

and v^2 = (2*9.8*2690) = 52724

The horizontal component of velocity is 131 m/s

The net magnitude of velocity

= [(Vertical component)^2 + (Horizontal component)^2]^(1/2)

= (52724 +131^2)^(1/2) = 264.357 m/s

tan of the angle this velocity makes with horizontal =

(vertical velocity)/(horizontal velocity) = 264.357/131 =  2.0180

For this value of tan, angle = 63.6 degrees

Answer: Overall velocity of hamper when it strikes the surface of the ocean is 264.357 m/s at 63.6 degrees to horizontal

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The height of the plane while droping the hamper = 2690m.

The hamper has two velocity components:(i) the horizontal velocity due its initial condition which is unaffected by gravity (ii) a vertical velocity which is increasing every moment due to gravity.

Since it is a drop, the vertical velocity of the hamper initially is zero. If the final vertical velocity while touching ocean is assumed v, then v^2-u^2=2gs, where u is initial velocity and v is the final velocity , g is the acceleration due to gravity and s is the traverse of vertical height.

v^2-0^2=2*9.8*2690. or v=sqrt(2*9.8*2690)=229.6171 m/s vertically down ward.

Therefore, the over all velocity is the resultant of the two velocities in two perpendicular directions i.e hprizontal and vertically downward directions= (horizontal velocity^2+vertically downward velocity ^2)^(1/2) = (131^2+229.6171^2)^(1/2) =264.3577m/s and the direction is declined to horizontal by an angle tan inverse (229.6171/131)=60.2946 degree.

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