A stone dropped from a balloon at rest from a certain height passes through 81 m in the last
before it strikes the ground. Find the height of the balloon and the velocity of the stone when it
hits the ground.
Answers
Answer:
A ball dropped from a balloon at rest clears a tower 81 m high during the last quarter second of its journey. What is the height of the balloon and the velocity of the body when it reaches the ground?
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2 Answers
Les McLean, Ph.D. in Engineering
Answered September 10, 2019 · Author has 4.7K answers and 3.8M answer views
Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):
s = ut + ½at^2 …. (1)
v^2 = u^2 + 2as …. (2)
v = u + at …. (3)
s = (u + v)t/2 …. (4)
where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.
In this case, we know from equation (1) that 81 = 0.25u + 4.905(0.25^2), so
u = 322.774m/s
Then from equation (4) 81 = (322.774 + v)0.25/2, or
v = 325.224m/s is the velocity at impact
Finally, from equation (2), 325.224^2 = 0 + 2(9.81)s, or
s = 5390.96m