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A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m>s. A 1.0-kg stone is thrown from the basket with an initial velocity of 15.0 m>s per- pendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m>s. (a) How high is the balloon when the rock is thrown

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Answer:

a) 242.5 m

Explanation:

Given :

Vi= 20m/s

t = 5 seconds

According to kinematic equation, the displacement is given by:

d = Vit + 1/2 gt^2

d = 20 × 6 ) + (1/2 × 9.8 × 5^2)

d = 120 + 122.5

d = 242.5m

Answer:

Explanation:

Given:

Mass = 124 kg

U = -20 m/s

t = 5 s

Using the equation of motion,

S = U×t + 1/2 × gt^2

S = -20 × 5 - 1/2 × 9.8 × 5^2

= -222.5 m

= 222.5 m (the sign is showing the direction of the motion from its origin)

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