Initial speed of Gazelle is along x direction and its value will be
[tex]v_x = 5.6 m/s[/tex]
also its initial height is given as
[tex]y = 2.5 m[/tex]
Part a)
now from kinematics along Y direction
[tex]\Delta y = v_y t + \frac{1}{2} at^2[/tex]
as we know that
[tex]\Delta y = 0[/tex]
[tex]v_y = 0[/tex]
[tex]a = 9.8 m/s^2[/tex]
[tex]2.5 = 0 + \frac{1}{2} (9.8) t^2[/tex]
[tex]t = 0.714 s[/tex]
Part b)
distance moved horizontally
[tex]\Delta x = v_x t[/tex]
as we know that
[tex]v_x = 5.6 m/s[/tex]
now we will have
[tex]v_x = 5.6 (0.714) = 4m[/tex]
so it will lend at distance of 4 m.
Part c)
final velocity in vertical direction
[tex]v_{fy} = v_y + at[/tex]
[tex]v_{fy} = 0 + (9.8)(0.714) = 7 m/s[/tex]
[tex]v_x = 5.6 m/s[/tex]
so net speed will be
[tex]v^2 = v_x^2 + v_y^2[/tex]
[tex]v^2 = 7^2 + 5.6^2[/tex]
[tex]v = 8.96 m/s[/tex]
