Respuesta :
Answer:
(a). The initial speed of the arrow is 49.96 m/s.
(b). The angle is 39.90°.
Explanation:
Given that,
Horizontal distance = 230 m
Time t = 6 sec
Vertical distance = 16 m
We need to calculate the horizontal component
Using formula of horizontal component
[tex]R =u\cos\theta t [/tex]
Put the value into the formula
[tex]\dfrac{230}{6} = u\cos\theta[/tex]
[tex]u\cos\theta=38.33[/tex].....(I)
We need to calculate the height
Using vertical component
[tex]H=u\sin\theta t-\dfrac{1}{2}gt^2[/tex]
Put the value in the equation
[tex]16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2[/tex]
[tex]u\sin\theta=\dfrac{16+9.8\times18}{6}[/tex]
[tex]u\sin\theta=32.06[/tex].....(II)
Dividing equation (II) and (I)
[tex]\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}[/tex]
[tex]\tan\theta=0.8364[/tex]
[tex]\theta=\tan^{-1}0.8364[/tex]
[tex]\theta=39.90^{\circ}[/tex]
(a). We need to calculate the initial speed
Using equation (I)
[tex]u\cos\theta\times t=38.33[/tex]
Put the value into the formula
[tex]u =\dfrac{230}{6\times\cos39.90}[/tex]
[tex]u=49.96\ m/s[/tex]
(b). We have already calculate the angle.
Hence, (a). The initial speed of the arrow is 49.96 m/s.
(b). The angle is 39.90°.
