A projectile is shot from the ground at an angle of #pi/4 # and a speed of #5 /8 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?

1 Answer
May 12, 2016

Let the velocity of projection of the object be u with angle of projection #alpha# with the horizontal direction.
The vertical component of the velocity of projection is #usinalpha# and the horizontal component is #ucosalpha#

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical dosplacement h will be zero. So applying the equation of motion under gravity we can write
#h=usinalpha xxT+1/2gT^2#
#=>0=uxxT-1/2xxgxxT^2#
,where #g="accelerationdueto""gravity”#
#:.T=(2usinalpha)/g#
The horizontal displacement during this T sec is #R=ucosalpha xxT#
# R=(u^2sin(2alpha))/g#
The time t to reach at the peak is half of time of flight (T)
So # t=1/2*T=(usinalpha)/g#
The horizontal displacement during time t is

#d_h=1/2xx(u^2sin(2alpha))/g#

By the question
#u =5/8m=5/8*100=62.5cm and alpha=pi/4#

#d_h=1/2xx((62.5)^2sin(2*pi/4)/980)=2cm#

If H is the maximum height then
# 0^2=u^2sin^2(alpha)-2*g*H#
#:. H= (u^2sin^2(alpha))/(2*g)#
#=((62.5)^2sin^2(pi/4))/(2*980)=1cm#

So the distance of the object from the point of projection when it is on the peak
Is given by
#D = sqrt(d_h^2+H^2)sqrt(2^2+1^2)=sqrt5=2.24cm#