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goldenfox [79]
4 years ago
12

A plane drops a hamper of medical supplies from a height of 5300 m during a practice run over the ocean. The plane is horizontal

velocity was 104 m/s at the instant the hamper was dropped. What is the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean
Physics
1 answer:
Pepsi [2]4 years ago
3 0

Answer:

v = 339 m/s

Explanation:

As we know that the horizontal speed of the plane is always constant

so here the speed in x direction always remain the same and given as

v_x = 104 m/s

now in order to find vertical speed we can use kinematics so we will have

v_f^2 - v_i^2 = 2a d

here we have

v_f^2 - 0^2 = 2(9.81)(5300)

v_y = 322.5 m/s

now net speed at which it will strike is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{104^2 + 322.5^2}

v = 339 m/s

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Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

v=\sqrt{v_x^2+v_y^2}           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

v_y^2=v_{oy}^2+2gh   (2)

vy: final velocity

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g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

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vx is calculated by using the information about the horizontal range of the ball:

R=v_o\sqrt{\frac{2h}{g}}    (3)

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You solve the previous equation for vo, the initial horizontal velocity:

v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}

The velocity of the ball just before it touches the ground is 46.99 m/s

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