Computing for angle and speed

Question

a horseand its rider jump over an obctcle 2.3m high while covering a horizontal distance of 10m. at what angle and with what speed did the horse leave the ground?

Answer 1

Let:

u be the the velocity required,

a be the inclination to the horizontal of its direction,

g be the acceleration due to gravity,

h be the height of the obstacle,

s be the horizontal distance of the obstacle,

t be the time taken to reach it.

Vertically:

0 = u^2 sin^2(a) - 2gh

u^2 sin^2(a) = 2gh ...(1)

0 = u sin(a) - gt

t = u sin(a) / g

Horizontally:

u cos(a) = s / t

= sg / [ u sin(a) ]

= sg / sqrt(2gh)

= s sqrt(g / [2h])

u^2 cos^2(a) = gs^2 / (2h) ...(2)

Adding (1) and (2):

u^2 = 2gh + gs^2 / (2h)

u^2 = g(4h^2 + s^2) / (2h) ...(3)

Dividing (1) by (2):

tan^2(a) = 4h^2 / s^2

tan(a) = 2h / s ...(4)

If 10m is the distance of the 2.3m high obstacle, then:

u = sqrt[ 9.81(4 * 2.3^2 + 100) / (2 * 2.3) ]

= 16.1 m/s.

a = arctan(2 * 2.3 / 10)

= 24.7 deg.