Question #8efea

2 Answers
Aug 3, 2016

Answer:

The ball is #(3h)/4# meters from the ground at #t = T/2#

Explanation:

I'm neglecting air resistance, but can add it in upon request. The body moves in only one dimension, which we shall call #y#. We shall define the origin of our coordinate system at the ground (#y=0#) and the initial position of the body #y(0) = h#. As is the usual convention we define upwards as positive.

Without air resistance, the only force acting on the body will be gravity - this acts in the negative y direction to provide an unbalanced force which we can use Newton's second law on:

#m(d^2y)/(dt^2) = -mg#

#(d^2y)/(dt^2)# represents acceleration in the y direction.

If you are unfamiliar with using calculus to solve problems like this then skip to the blue heading, this is just the derivation of a formula common to introductory dynamics.

#therefore (d^2y)/(dt^2) = -g#

Integrating wrt t gives:

#(dy)/(dt) = -g t + C#

The body starts from rest so #y'(0) = 0#.

#implies C =0#

#(dy)/(dt) = -g t#

Integrate again wrt t:

#y(t) = -1/2g t^2 + C#

#y(0)= h = C#

#implies y(t) = h - 1/2g t^2#

This is our expression for the displacement from the origin as a function of time. Notice it's very similar to the common expression

#s = v_0t + 1/2at^2#

which describes displacement under constant acceleration. This is because it is the exact same derivation, we have just accounted for the fact that initial speed is zero while doing it.

#color(blue)("Skip To Here")#

Anyway, onto solving the problem:

#y(t) = h - 1/2g t^2#

If it takes T seconds to reach the ground, #y(T) = 0#.

#y(T) = 0 = h - 1/2gT^2#

#implies h = 1/2gT^2#

#T^2 = (2h)/g#

#T = sqrt((2h)/g)#

Note that we discard the negative solution from the square root because negative time doesn't really make sense.

#T/2 = 1/2sqrt((2h)/g) = sqrt(h/(2g))#

#y(T/2) = h - 1/2g (sqrt(h/(2g)))^2#

#y(T/2) = h - 1/2g(h/(2g)) = h - h/4 = (3h)/4#

Aug 4, 2016

Answer:

The body is at #h/4 m# from the top of tower.

Explanation:

Kinematic equation for the body dropped from a height #h# is
#h=ut+1/2g t^2#
where #u,g and t# are initial velocity, acceleration due to gravity and time taken to reach ground respectively.
Inserting given values we get
#h=0xxT+1/2g T^2#
#=>h=1/2g T^2# .......(1)

Similarly height #h_1# dropped in time #T/2# is
#h_1=1/2g (T/2)^2#
#=>h_1=1/8g T^2#......(2)
Dividing (2) by (1) we get
#h_1/h=1/4#

#=>h_1=h/4#