Answer:[tex]u=4.51 m/s\ at\ angle\ of\ \theta =59.34^{\circ}[/tex]
Explanation:
Given
height of building h=3 m
Landing velocity of diver [tex]v=8.90 m/s[/tex] at an angle of [tex]75^{\circ}[/tex]
Let u be the initial velocity of diver at an angle of \theta with horizontal
Since there is no acceleration in horizontal direction therefore horizontal component of velocity will remain same
[tex]u\cos \theta =8.9\cos (75)[/tex] ---- -----1
Considering Vertical motion
[tex]v^2-u^2=2as[/tex]
here [tex]v=8.9\sin (75)[/tex]
[tex]u=u\sin \theta[/tex]
[tex]s=3 m[/tex]
[tex]a=9.8 m/s^2[/tex]
[tex](8.9\sin (75))^2-(u\sin \theta )^2=2\times 9.8\times 3[/tex]
[tex]u\sin \theta =\sqrt{(8.9\sin 75)^2-(2\cdot 9.8\cdot 3)}[/tex]
[tex]u\sin \theta =3.886[/tex] ----------------2
Divide 2 and 1 we get
[tex]\tan \theta =\frac{3.886}{8.9\cos (75)}[/tex]
[tex]\tan \theta =1.687[/tex]
[tex]\theta =59.34^{\circ}[/tex]
Thus [tex]u\cos (59.34)=8.9\cos (75)[/tex]
[tex]u=4.51 m/s[/tex]
