A projectile is shot from the ground at an angle of #pi/4 # and a speed of #12 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

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Answer 1 · 2016-06-11T11:27:01

#d=8.2141272642757" "#meters

Given #theta=pi/4#
initial velocity #v_0=12" "#m/sec

Compute the total time t it will be on the air

#y=v_0*sin theta*t+1/2*g*t^2#

#0=12*sin (pi/4)*t+1/2(-9.8)*t^2#

#0=12*(1/sqrt2)*t-4.9*t^2#

#0=6sqrt2*t-4.9*t^2#

#0=t(6sqrt2-4.9t)#

#t_1=0# and #t_2=(6sqrt2)/4.9#

Let #x=#range

We will use #t_2=(6sqrt2)/4.9#

#x=v_0 cos theta*t#

#x=12*cos (pi/4)(6sqrt2)/4.9#

#x=14.69387755" "#meters

We need the half-range #x/2=7.3469387755" "#meters

We also need the maximum height #y_max#

#y=v_0*sin thetat+1/2g*t^2#

At #t_2/2# the projectile is at its maximum height.

#y_max=v_0*sin theta*t_2/2+1/2g*(t_2/2)^2#

#y_max=12sin (pi/4)((6sqrt2)/4.9)/2+1/2(-9.8)(((6sqrt2)/4.9)/2)^2#

#y_max=3.673469387755" "#meters

Distance #d# of the projectile from the starting point

#d=sqrt((x/2)^2+y_max^2)#

#d=sqrt((7.3469387755)^2+3.673469387755^2)#

#d=8.2141272642757" "#meters

God bless....I hope the explanation is useful.