A particle is moving vertically upward and reaches the maximum height H in T seconds. The height of the particle at any time t ( t > T) will be?

A: #H - g( t - T )^2#
B: #g(t - T)^2#
C: #H - 1/2g(t - T)^2#
D: #g/2(t - T)^2#

1 Answer
Nathan L.
Aug 4, 2017

(C) #"height" = H - 1/2g(t-T)^2#

Explanation:

When the particle is at its maximum height, its motion after this point is analogous to that of a particle dropped from rest at a certain height #H# and time #T#.

With that being said, we can use the kinematics equation

#ul(y = y_0 + v_(0y)t - 1/2g t^2#

where

  • the initial height #y_0# is the maximum height #H#

  • the initial velocity #v_(0y)# is #0# (equivalent to being dropped from rest)

  • #t# represents all times after #T#, i.e. is equivalent to the expression #t-T#

Since the initial velocity is #0#, that leaves us with

#ul(y = y_0 - 1/2g t^2#

Plugging in values from above yields

#color(red)(ulbar(|stackrel(" ")(" "y = H - 1/2g(t-T)^2" ")|)#

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