Answer:
time is 32 s and speed is 304.3 m/s
Explanation:
Height, h = 146 m
speed, u = 14 m/s
Angle, A = 43 degree
Let it hits the ground after time t.
Use second equation of motion
[tex]h = u t +0.5 at^2\\\\- 146 =14 sin 43 t - 4.9 t^2\\\\4.9 t^2 - 9.5 t - 146 =0 \\\\t =\frac{9.5\pm\sqrt {90.25 + 2861.6}}{9.8}\\\\t=\frac{9.5\pm 54.3}{9.8}\\\\t = 32.05 s, - 22.4 s[/tex]
Time cannot be negative so the time is t = 32 s .
The vertical velocity at the time of strike is
v' = u sin A - g t
v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s
horizontal velocity
v'' = 14 cos 43 =10.3 m/s
The resultant velocity at the time of strike is
[tex]v=\sqrt{v'^2 + v''^2}\\\\v = \sqrt{304.1^2 +10.3^2 }\\\\v = 304.3 m/s[/tex]
