How do a calcuate this projectile?

Question

A projectile is launched from a point at a horizontal distance d from the base of a vertical wall of height h. Show that the minimum speed of projection to enable the projectile to pass over the wall is

[g(h + √h^2 + d^2)]^1/2.

Any help would be greatly appreciated.

Answer 1

Let:

u be the speed of projection,

t be the time to reach the wall,

a be the angle of projection measured from the horizontal.

Horizontally:

d = ut cos(a) ...(1)

Vertically:

h = ut sin(a) - gt^2 / 2 ...(2)

Substituting for t from (1) in (2):

h = d sin(a) / cos(a) - gd^2 / [ 2u^2 cos^2(a) ]

2hu^2 cos^2(a) = 2du^2 sin(a)cos(a) - gd^2

u^2 [ 2d sin(a)cos(a) - 2h cos^2(a) ] = gd^2

Using

sin(2a) = 2 sin(a) cos(a)

cos(2a) = 2 cos^2(a) - 1:

u^2 [ d sin(2a) - h( 1 + cos(2a) ) ] = gd^2 ...(3)

2u (du / da) [  d sin(2a) - h( 1 + cos(2a) ) ] + u^2 [ 2d cos(2a) + 2h sin(2a) ] = 0

du / da = - u^2 [ 2d cos(2a) + 2h sin(2a) ] / { 2u [  d sin(2a) - h( 1 + cos(2a) ) ] }

du / da = 0 when:

d cos(2a) = - h sin(2a)

tan(2a) = - d / h

2a is a 2rd quadrant angle.

sin(2a) = d / sqrt(h^2 + d^2) ...(4)

cos(2a) = - h / sqrt(h^2 + d^2) ...(5)

Substituting from (4) and (5) in (3):

u^2[ d^2 / sqrt(h^2 + d^2) - h( 1 - h / sqrt(h^2 + d^2) ) ] = gd^2

u^2[ d^2 - h sqrt(h^2 + d^2) + h^2 ] = gd^2 sqrt(h^2 + d^2)

u^2 = gd^2 sqrt(h^2 + d^2) / [ h^2 + d^2 - h sqrt(h^2 + d^2) ]

= gd^2 sqrt(h^2 + d^2) [ h^2 + d^2 + h sqrt(h^2 + d^2) ] / [ (h^2 + d^2)^2 - h^2(h^2 + d^2) ]

= gd^2 sqrt(h^2 + d^2) [ h^2 + d^2 + h sqrt(h^2 + d^2) ] / [ d^2(h^2 + d^2) ]

= g sqrt(h^2 + d^2)[ h^2 + d^2 + h sqrt(h^2 + d^2) ] / (h^2 + d^2)

= g[ h^2 + d^2 + h sqrt(h^2 + d^2) ] / sqrt(h^2 + d^2)

u^2 = g[ h + sqrt(h^2 + d^2) ].