Answer:
16.17 m/s
Explanation:
h = 3.2 m
u = 18.1 m/s
Angle of projection, θ = 49°
Let H be the maximum height reached by the ball.
The formula for the maximum height is given by
[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]
[tex]H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m[/tex]
The vertical distance fall down by the ball, h' H - h = 9.52 - 3.2 = 6.32 m
Let v be the velocity of ball with which it strikes the ground.
Use third equation of motion for vertical direction
[tex]v_{y}^{2}=u_{y}^{2}+2gh'[/tex]
here, uy = 0
So,
[tex]v_{y}^{2}=2\times 9.8 \times 6.32[/tex]
vy = 11.13 m/s
vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s
The resultant velocity is given by
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]
[tex]v=\sqrt{11.87^{2}+11.13^{2}}[/tex]
v = 16.27 m/s
