Respuesta :
Answer:
Explanation:
Given
launch angle [tex]\theta _1=30^{\circ}[/tex]
[tex]\theta _2=60^{\circ}[/tex]
[tex]\theta _1[/tex] and [tex]\theta _2[/tex] are complimentary angles so range for both of them is same
[tex]H_{max}=\frac{u^2(\sin \theta )^2}{2g}[/tex]
[tex]H_{30}=\frac{u^2(\sin 30)^2}{2g}[/tex]
[tex]H_{30}=\frac{u^2}{8g}[/tex]
time of flight [tex]=\frac{2u\sin \theta }{g}[/tex]
[tex]t_{30}=\frac{u}{g}[/tex]
For [tex]\theta =60^{\circ}[/tex]
[tex]H_{60}=\frac{u^2(\sin 60)^2}{2g}[/tex]
[tex]H_{60}=\frac{3u^2}{8g}[/tex]
[tex]t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}[/tex]
[tex]H_{60}=3\times H_{30}[/tex]
[tex]t_{60}=\sqrt{3}\times t_{30}[/tex]
The projectile launched at 60° attained more height and also spent more time in air when compared to the projectile launched at 30° to the horizontal.
The given parameters;
- angle of projection of the first projectile, θâ = 30â°
- angle of projection of the second projectile, θâ = 60â°
The maximum height reached by each projectile is calculated as follows;
[tex]H= \frac{u^2 (sin\ \theta)^2}{2g} \\\\H_{30} = \frac{u^2 (sin \ 30)^2}{2\times 9.8}\\\\H_{30} = 0.0128 \ u^2[/tex]
[tex]H= \frac{u^2 (sin\ \theta)^2}{2g} \\\\H_{60} = \frac{u^2 (sin \ 60)^2}{2\times 9.8}\\\\H_{60} = 0.0383 \ u^2[/tex]
The time of flight of the two projectiles is calculated as follows;
[tex]T = \frac{2u\times sin(\theta)}{g} \\\\T_{30} = \frac{2u \times sin( 30)}{9.8} \\\\T_{30} = 0.102 \ u[/tex]
[tex]T_{60} = \frac{2u \times sin( 60)}{9.8} \\\\T_{60} = 0.177 \ u[/tex]
Thus, the projectile launched at 60° attained more height and also spent more time in air when compared to the projectile launched at 30° to the horizontal.
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