Question

8) A ball dropped from a balloon at rest clears a
tower 81 m high during the last quarter second
of its journey. Find the height of the balloon
and the velocity of the body when it reaches
the ground​

Answer 1

Answer:

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

The balloon was at a height of 5390.96m

Explanation:

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