Question

A gray kangaroo can bound across level ground with each jump carrying it 9.9 m$\mathrm{m}$ from the takeoff point. Typically the kangaroo leaves the ground at a 30 ∘${}^{\circ }$ angle. If this is so: what is it's takeoff speed and maximum height above the ground?

Answer 1

The launch speed of the kangaroo can be calculated by using the following equation. v2 sin 29 Here, R is the range, θ launch angle, g is acceleration due to gravity is launch speed, and 82m - lö'sin 2(21°) 9.8 m/s v, = 10.96 m/s The maximum height reached by the kangaroo can be calculated by using the following equation (10.96 m/s)sin 21)-(0) 2(9.8 m/s* 0.787 m Similarly, calculate the launch speed and the maximum height reached by the kangaroo for the other values The launch speed of the kangaroo can be calculated by using the following equation. v2 sin 29 Here, R is the range, θ launch angle, is launch speed, and g is acceleration due to gravity 9.2 m v10.7 m/s The maximum height reached by the kangaroo can be calculated by using the following o sin 2( 26 9.8 m/s equation (10.7 m/s)sin 26)-(0)* 2(9.8 m/s 1.12 m