Question

PLease help!!!!!!
A scared kangaroo once cleared a fence that was 2.44m high. If the Horizontal component of the kangaroo’s velocity was 4.80 m/s, find the angle with respect to the ground at which the kangaroo leaped.

Answer 1

For this type of question, you'll have to find the time first.
Since \(S_y=v_y t - \frac{1}{2} a_y t^2\), where \(S_y =2.44, v_y=0, a_y=-9.8\), you'll get t.
Now, the next thing is get \( u_y\).
\(S_y=u_y t + \frac{1}{2} a_y t^2\). Now sub all the things back to get \(u_y\)
After that, since \(u_x = u \cos \theta\) and \(u_y = u \sin \theta\)
\(\frac{u_y}{u_x} = \tan \theta\)
So, \(\theta= \tan^{-1}\frac{u_y}{u_x}\) Sub it all in and voila! :)
Do you need further help?

Answer 2

Initial vertical velocity must be different of zero, in order the Kangaroo jump the fence
\[V _{oy}\neq0\] the wall height is h= 2.44 m \[V _{fy}^{2} - V _{oy}^{2} = 2 g h\] final kangaroo vertical velocity is zero. \[V _{fy} = 0\] initial vertical velocity is: \[V _{oy} = \sqrt{2 g h} = 6.91 m/s\]
Horizontal velocity is: \[V _{x} = 4.80 m/s\]
\[\tan (alfa) = V _{0y} / V _{x} = 6.91/4.80 = 0.963673\]
\[alfa = 0.963673 \times 180 \div \pi = 55.21 grados \]