Answer:
a) u = 9.88 m/s
b) [tex]\theta = 79^0[/tex]
Explanation:
given,
speed of the diver = 13.2 m/s
Angle made w.r.t horizontal = 81.8 °
The horizontal component of final velocity
v_h = 13.2 cos 81.8 °
The vertical component of final velocity
v_v = 13.2 sin 81.8 °
using equation of motion
initial velocity of vertical component
[tex]u_v = \sqrt{v_v^2-2as}[/tex]
[tex]u_v = \sqrt{(13.2 sin 81.8^0)^2-2(-9.8)(-3.9)}[/tex]
= 9.70 m/s
the horizontal component of the initial velocity
u_h = v_h = 13.2 cos 81.8 ° = 1.88 m/s
magnitude
[tex]u = \sqrt{9.7^2+1.88^2}[/tex]
u = 9.88 m/s
direction
[tex]\theta = tan{-1}(\dfrac{9.7}{1.88})[/tex]
[tex]\theta = 79^0[/tex]
