A man stands on a crane and throws a water balloon down at at 21 m/s. He finds that it takes 2.4s for the balloon to hit the ground. What is its height one second before it hits the ground?

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Answer 1 · 2015-09-01T12:17:53

#39.62" m"#

Consider the Cartesian frame reference. Meaning that the downward direction is taken to be negative and the upward direction is positive

#H# stands for the maximum height
#h# is the height at any time #t# during the balloon's fall

From newton's equation of motion for displacement:
#y=ut+1/2at^2#

In the present situation,

#y# is the displacement
Considering the fact that, height #h# at any time can be gotten from the equation: #h= H+y#

Thus, #y=h-H#

#t# is time of fall,
#t= 2.4-1= 1.4#

#a# is acceleration
#a= -g= -9.8 ms^-2#

#u# is initial velocity or velocity at time #t=0#
#u=-21ms^-1#

The negative signs means that they are directed downwards!

Substituting all the variables into #y=ut+1/2at^2# , we get:

#=>h-H=(-21)(1.4)+1/2(-9.8)(1.4)^2=-39.004#

#=>h=H-39.004#

To find #H# we consider the case where the balloon goes all the way down and hits the ground

This time,
#t=2.4s# and #y=-H#
The other data remain the unchanged.

#-H=(-21)(2.4)+1/2(-9.8)(2.4)^2=-78.624#

#=>H=78.624#

#=>h=78.624-39.004=color(blue)(39.62" m")#