Question #14c89

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Answer 1 · 2017-09-20T06:53:08

The average speed is #=(10)ms^-1#

Assume that the height of the tree is #=hm=5m#

Resolving in the direction #darr_+#

The initial speed of the apple is #u=0ms^-1#

Apply the equation of motion

#s=ut+1/2at^2#

So,

#h=0+1/2g t^2#

Where #t# is the time to reach the ground

Therefore,

#t^2=(2h)/g#, #<=>#, #t=sqrt((2h)/g)#

The distance travelled by Bob is #d=50m#

Bob's average speed is

#barv=d/t=50/sqrt((2h)/g)=(50sqrt5)/(sqrth)ms^-1#

#=(50sqrt5)/(sqrt5)ms^-1#

#=50ms^-1#