Respuesta :
Answer:
5.79 m/s
Explanation:
R=2.9 m
H=0.436 m
∅=44.7°
For Horizontal Motion
aₓ=0 uₓ=ucos∅
x=uₓt+[tex]\frac{1axt^{2} }{2}[/tex]
R=ucos∅*t+0
2.9=ucos(38.3)*t
t=[tex]\frac{4.079}{u}[/tex]
For Vertical Motion
Y=Uy*t+[tex]\frac{1ayt^{2} }{2}[/tex]
By putting value
0.436=usin∅([tex]\frac{4.079}{u}[/tex])-([tex]\frac{1}{2}g *(\frac{4.079}{u} )^{2}[/tex])
0.436=sin(44.7)([tex]\frac{4.079}[/tex])-([tex]\frac{1}{2}g *(\frac{4.079}{u} )^{2}[/tex])
0.436=2.869-[tex]\frac{81.61}{u^{2} }[/tex]
u²=33.54
u=5.79 m/s
Answer:
5.83 m/s
Explanation:
This is a projectile motion problem.
Let's call
[tex]V_0 [/tex] the initial velocity
In x-coordinate:
[tex]V_{0x} = V_0 \times cos \alpha [/tex]
where α is 44.7°
In y-coordinate:
[tex]V_{0y} = V_0 \times sin \alpha[/tex]
In x-coordinate the displacement is:
[tex]x = V_0 \times cos \alpha \times t [/tex]
where t is time. Isolating the initial speed and replacing with x = 2.9 (the distance travelled in x direction) :
[tex] V_0 = \frac{2.9}{cos \alpha \times t} [/tex]
In y-coordinate the displacement is:
[tex]y = V_0 \times sin \alpha \times t - 1/2 \times g \times t^2[/tex]
Replacing with the initial velocity, y = 0.436, alpha = 44.7° and g = 9.81:
[tex]0.436 = \frac{2.9}{cos \alpha \times t} \times sin \alpha \times t - 1/2 \times g \times t^2[/tex]
[tex]0.436 =2.9 \times tan 44.7 - 1/2 \times 9.81 \times t^2[/tex]
[tex] t^2 = \frac{2.9 \times tan 44.7 - 0.436}{ 1/2 \times 9.81} [/tex]
[tex]t = \sqrt{0.495}[/tex]
[tex]t = 0.7[/tex]
Replacing this value in the previous initial velocity equation:
[tex] V_0 = \frac{2.9}{cos 44.7 \times 0.7} [/tex]
[tex] V_0 = 5.83 \; m/s [/tex]
