A projectile is shot from the ground at an angle of #pi/6 # and a speed of #3 m/s#. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

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Answer 1 · 2016-04-24T12:19:00

#y_max=+0.114795" "#meter#=11.4795" "#centimeter
#x=+0.3976647262" "#meter#=39.76647262" "#centimeter

Equations for projectiles are as follows

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

Time from the ground to its maximum height
#t=(-v_0 sin theta)/g#

Equation for the maximum height #y_max#

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

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

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

#y_max=(-v_0^2*sin^2theta)/(2g)#

#y_max=(-(3)^2*(sin 30^@)^2)/(2*(-9.8))#

#y_max=+0.114795" "# meter

Horizontal distance

#x=v_0*cos theta*t#

#x=3*cos 30^@ ((-v_0 sin theta)/g)#

#x=3*cos 30^@ *((-3*sin(30^@))/(-9.8))#

#x=+0.3976647262" "#meter

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