Question #dc83d

2 Answers
Jun 1, 2016

Answer:

#E_p/E_k=sec^2 alpha-1#

Explanation:

#v=v_x=v_i*cos alpha" (Horizontal component of velocity of object)"#

#E_k=1/2*m*v_x^2#

#E_k=1/2*m*v_i^2*cos^2 alpha#

#h_m=v_i^2*sin^2 alpha/(2*g)" maximum height of object"#

#E_p=m*g*h_m#

#E_p=m*cancel(g)*(v_i^2*sin^2 alpha)/(2*cancel(g))#

#E_p=(m v_i^2*sin^2 alpha)/2#

#E_p/E_k=(cancel(1/2*m)*cancel(v_i^2)*sin^2 alpha)/(cancel(1/2*m) cancel(v_i)^2*cos^2alpha)#

#E_p/E_k=sin^2 alpha/cos^2 alpha#

#sin^2 alpha=1-cos^2 alpha#

#E_p/E_k=(1-cos^2 alpha)/(cos^2 alpha)#

#E_p/E_k=(1/cos^2 alpha)-1#

#1/cos^2 alpha=sec^2 alpha#

#E_p/E_k=sec^2 alpha-1#

Jan 21, 2017

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As referred to the figure above.

Maximum height of the object having mass #m# will be reached when #y#- or #sin theta# component of projectile's velocity becomes #0# (zero).

As such kinetic energy at this point is given by
#KE_"max height"=1/2m(ucos theta)^2#

At this point all the kinetic energy due to #y#-component of initial velocity has got converted into its Potential Energy which is also given as #mgh#
#PE_"max height"=mgh=1/2m(usin theta)^2#

Ratio of the two is #(PE_"max height")/(KE_"max height")=(1/2m u^2sin^2 theta)/(1/2m u^2cos^2theta)#
#=tan^2 theta#