A small mailbag is released from a helicopter that is descending steadily at 2.9

Question

A small mailbag is released from a helicopter that is descending steadily at 2.90 m/s. (a) After 4.00 s, what is the speed of the mailbag? v = m/s (b) How far is it below the helicopter? d = m (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.90 m/s? v = m/s d = m A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 2.80 s for the ball to reach its maximum height. (a) Find the ball's initial velocity. m/s upward b) Find the height it reaches

Explanation / Answer

a) initial velocity= -2.9 m/s as coming down

a=-9.8 m/s2

t= 4 s

v=u+at

v=-2.9-9.8*4 = -42.1 m/s , Seepd = 42.1 m/s

b) s= ut+0.5 a t^2

-2.9*4-0.5*9.8*4^2 = -90 m so the report as 90 m

c) If its going up then use u=+2.9 m/s in the above equation

v=2.9-9.8*4= -36.3 m/s , speed=36.3 m/s

s=2.9*4-0.5*9.8*4^2 = -66.8 m , report 66.8 m

second question

v=u+at ( at the max height v=o)

0=u-9.8*2.8

u=27.44 m/s (Initial velocity)

Find the height it reaches

v^2=u^2+2 a s

0=27.44^2-1*9.8*s

s=38.416 m

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