Ask Question
25 June, 17:08

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

+2
Answers (2)
  1. 25 June, 19:29
    0
    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
  2. 25 June, 20:40
    0
    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)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers