Answer:
7.84 m/s
Explanation:
Height, h = 2 m
Initial velocity, u = 10 m/s
Angle, A = 80°
(a) Let the time taken to go to the net is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]
Time cannot be negative.
So, t = 0.2 s
The vertical velocity at t = 0.2 s is
v = u + at
v = 10 sin 80 - 9.8 x0.2
v = 9.8 - 1.96 = 7.84 m/s
