A projectile is shot from the ground at an angle of #pi/8 # and a speed of #2 m/s#. When the projectile is at its maximum height, what will its distance from the starting point be?

2 Answers
mason m
Nov 23, 2016

#t_m=(2sin(pi/8))/g#

#d=(2sin(pi/8)(4cos^2(pi/8)+sin^2(pi/8)))/g#

Explanation:

#a_y(t)=-g#.

#v_y(t)=inta_y(t)dt=-g t+v_(0y)=-g t+v_0sin(theta)#.

#y(t)=intv_y(t)=-1/2g t^2+v_0sin(theta)t#.

#a_x(t)=0#.

#v_x(t)=inta_x(t)=0+v_(0x)=v_0cos(theta)#.

#x(t)=intv_x(t)=v_0cos(theta)t#.

The projectile is at its maximum height when #v_y(t)=0#, so when #g t=v_0sin(theta)#.

#t_m=(v_0sin(theta))/g=(2sin(pi/8))/g#.

Its #x# distance from the starting point at this time will be #x(t_m)=(v_0cos(theta)*v_0sin(theta))/g=(4sin(pi/8)cos(pi/8))/g#.

Its #y# distance from the starting point at this time will be #y(t_m)=-1/2g((v_0sin(theta))/g)^2+v_0sin(theta)((v_0sin(theta))/g)=(v_0^2sin^2(theta))/(2g)=(2sin^2(pi/8))/g#.

So the total distance #d# from the starting point is given by the Pythagorean Theorem on the distances, i.e. #d^2=x^2+y^2#

So after all the algebra, we get #d=(2sin(pi/8)(4cos^2(pi/8)+sin^2(pi/8)))/g#.

Douglas K.
Nov 23, 2016

The distance form the starting point, #d ~~ 0.15 m#

Explanation:

The equation for the y coordinate is:

#y(t) = ((-4.9" m")/"s"^2)t^2 + ((2" m")/s")sin(pi/8)t#

The equation for the x coordinate is:

#x(t) = ((2" m")/s")cos(pi/8)t#

The time that the projectile is at is maximum height is at the axis of symmetry of the parabola:

#t = -b/(2a)#

#t = -(2sin(pi/8))/(-9.8)s#

#t ~~ 0.08" s"#

#y(0.08) ~~ 0.03 m#

#x(0.08) ~~ 0.15m#

The distance, d, at the above time is:

#d = sqrt((0.03)^2 + (0.15)^2)#

#d ~~ 0.15 m#

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