A projectile is shot from the ground at an angle of #pi/6 # and a speed of #4 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 · 2016-11-25T11:19:49

The maximum height is #=0.2m#

The horizontal distance is #=0.69m#

The maximum height is when the vertical component of the velocity #=0#

We use the equation

#v^2=u^2-2gy#

#0=(v_0sintheta)^2-2gy#

#y=(4*sin(pi/6))^2/(2g)#

#y_(max)=2/g=2/10=0.2m#

The time to reach the greatest height is #t=v_0sintheta/g#

#t=4*sin (pi/6)/10=0.2s#

The horizontal distance is #x=v_0costheta*t#

#=4*cos(pi/6)*0.2=0.8*sqrt3/2=0.4sqrt3=0.69m#

graph{y=(-5x^2/12)+0.58x [-0.819, 1.881, -0.032, 1.318]}