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A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 5 m/s. The bag is released with the same upward velocity of 5 m/s when t = 0 and hits the ground when t = 8 s.

(a) Determine the speed of the bag as it hits the ground.

(b) Determine the altitude of the balloon when the bag hits the ground

Respuesta :

Answer:

a) 73.48 m/s

b) 313.92 m

Explanation:

Data provided in the question:

ascending velocity = - 5 m/s      [ negative sign depicts upward movement]

Time taken by bag to hit the ground, t = 8 s

a) from the Newton's equation of motion

we have  

[tex]s=ut+\frac{1}{2}at^2[/tex]

where,  

u is the initial speed  

a is the acceleration = 9.81 m/s²   (since it is a case of free fall )

s is the distance

thus,

[tex]s=(-5)(8)+\frac{1}{2}(9.81)(8)^2[/tex]

s = - 40 + 313.92

s = 273.92 m

from

v = u + at

v is the final speed

v = -5 + (9.81)(8)

or

v = 73.48 m/s

b) Distance traveled by balloon  = Speed × Time

= 5 × 8

= 40 m

Therefore,

Altitude of the balloon

= Distance traveled by bag + Distance traveled by balloon

= 273.92 m + 40 m

= 313.92 m

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