Answer:
(a) 38.5m/s
(b) 64.4m/s
Explanation:
First, we can obtain the launch speed from the definition of kinetic energy:
[tex]K=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2K}{m}}\\\\[/tex]
Plugging in the given values, we obtain:
[tex]v=\sqrt{\frac{2(1550J)}{0.55kg}}\\\\v=75.0m/s[/tex]
Now, from the conservation of mechanical energy, considering the instant of launch and the instant of maximum height, we get:
[tex]E_0=E_f\\\\K_0=U_g_f+K_f\\\\\frac{1}{2}mv_0^2=mgh_f+\frac{1}{2}mv_0_x^2\\\\\frac{1}{2}mv_0^2=mgh_f+\frac{1}{2}mv_0^2\cos^2\theta\\\\\implies \cos\theta=\sqrt{1-\frac{2gh_f}{v_0^2}}[/tex]
And with the known values, we compute:
[tex]\cos\theta=\sqrt{1-\frac{2(9.8m/s^2)(140m)}{(75.0m/s)^2}}\\\\\cos\theta=0.513\\\\\theta=59.12\°[/tex]
Finally, to know the components of the launch velocity, we use trigonometry:
[tex]v_0_x=v_0\cos\theta=(75.0m/s)\cos(59.12\°)=38.5m/s\\\\v_0_y=v_0\sin\theta=(75.0m/s)\sin(59.12\°)=64.4m/s[/tex]
It means that the horizontal component of the launch velocity is 38.5m/s (a) and the vertical component is 64.4m/s (b).
