Help me find the max horizontal distance ?

Question

A kangaroo is capable of bouncing to a max horizontal distance of 8 m.

what is the speed of his jump?

please explain yourself when solving - thx!

Answer 1

Sorry, unable to solve for distance over time (speed) knowing only distance (8m).

However, assuming a launch angle for the jump of 45 degrees and treating the kangaroo as a projectile with no air resistance:

8 m is then the range of the jump.

R = v^2 * (sin 2*θ) / g

v^2 = R * g / (sin 2*θ)

v = √ (R * g / (sin 2*θ))

Assuming the launch angle is 45 degrees (yielding the longest possible jump)

R = range of jump = 8 m

θ = 45°

g = 9.8 m/s/s

v = initial velocity of the jump

v = √(8 * 9.8 /(sin 2*45) = √ ( 78.4 / 1 ) = 8.85 m/s

the initial speed of a kangaroo jumping 8 m at an angle of 45° with no air resistance.