Respuesta :
Answer:[tex]60^{\circ}[/tex]
Explanation:
Given
speed of ball [tex]u=5 m/s[/tex]
launch angle [tex]\theta =30^{\circ}[/tex]
Range of a Projectile [tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
Range will be common for two angles i.e. [tex]\theta [tex] and [tex]90-\theta [/tex]
for [tex]\theta R=\frac{5^2\sin 2\cdot 30}{g}[/tex]
[tex]R=2.209 m[/tex]
For [tex]90-\theta [/tex]
[tex]90-30=60^{\circ}[/tex]
[tex]R=\frac{5^2\sin 2\cdot 60}{g}[/tex]
[tex]R=2.209 m[/tex]
Answer:
The other angle is 60°
(a) is correct option.
Explanation:
Given that,
Angle = 30
Speed = 5.0 m/s
We need to calculate the range
Using formula of range
[tex]R=\dfrac{v^2\sin(2\theta)}{g}[/tex]...(I)
The range for other angle is
[tex]R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}[/tex]...(II)
Here, distance and speed are same
From equation (I) and (II)
[tex]\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}[/tex]
[tex]\sin(2\theta)=\sin(2(\alpha-\theta))[/tex]
Put the value into the formula
[tex]\sin(2\times30)=\sin(2\times(\alpha-\theta))[/tex]
[tex]\sin60=\sin2(\alpha-30)[/tex]
[tex]60=2\alpha-60[/tex]
[tex]\alpha=\dfrac{60+60}{2}[/tex]
[tex]\alpha=60^{\circ}[/tex]
Hence, The other angle is 60°.
