A projectile is shot from the ground at an angle of #pi/4 # and a speed of #15 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?

2 Answers
May 5, 2017

The distance is #=11.5m#

Explanation:

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=15*sin(1/4pi)#

Acceleration is #a=-g#

At the maximum height, #v=0#

We apply the equation of motion

#v=u+at#

to calculate the time to reach the greatest height

#0=15sin(1/4pi)-g*t#

#t=15/g*sin(1/4pi)#

#=1.08s#

Resolving in the horizontal direction #rarr^+#

To find the horizontal distance, we apply the equation of motion

#s=u_x*t#

#=15cos(1/4pi)*1.08#

#=11.5m#

May 5, 2017

enter image source here
Here in our case
#theta=pi/4#
#u=15#m/s
#:.u_(y_1)=15sin(pi/4)=15/sqrt2#
#u_x=15cos (pi/4)=15/sqrt2#
When the projectile will reach maximum height
Velocity in #y# direction #u_(y_2)##=0# we
#:.# Applying equating of motion along #y # direction we get
#(u_(y_2))^2 = (u_(y_1))^2 +2 (-g)h#
From this equation we get #h=5.74m#
Now time required to reach the highest point
#t=u_(y_2)/g#
#:.# distance moved in #x# ditection in this time
#d=u_(x)##xxu_(y_2)/g#
#:.d=#11.5#m#
#:. #Distance from starting point of the projectile till it reaches the maximum height is given by

#l=sqrt (d^2+h^2)#
#:.l=12.8m#