Respuesta :
Question :-
- The maximum vertical height of a projectile is 10 m. If the magnitude of the initial velocity is 28 m/s, what is the direction of the initial velocity? (g=9.8 m/s2 )
Required Solution :-
- The direction of the initial velocity is 30° from the horizontal.
Provided that:-
- Maximum vertical height = 10 m
- Magnitude of initial velocity = 28 m/s
To Calculate:-
- Direction of initial velocity
Using Formula :-
- [tex]\bigstar \red{\underline{\boxed{\pmb{\sf\bf\red{h = \dfrac{u^2 sin^2 \theta}{2g}}}}}}[/tex]
Calculation :-
[tex] \qquad \pmb{\sf\longrightarrow 10 = \dfrac{(28)^2 \times sin^2 \theta}{2 \times 9.8}}[/tex]
[tex] \qquad \pmb {\sf \longrightarrow 10 = \dfrac{784 \times sin^2 \theta}{19.6}}[/tex]
[tex] \qquad \pmb{\sf \longrightarrow 196 = 784 \times sin^2 \theta}[/tex]
[tex]\qquad \pmb{\sf \longrightarrow sin^2 \theta = \dfrac{196}{784}}[/tex]
[tex]\qquad \pmb{\sf \longrightarrow sin^2 \theta = 0.25 }[/tex]
[tex]\qquad \pmb{\sf\longrightarrow sin \theta = \sqrt{0.25}}[/tex]
[tex]\qquad \pmb{\sf\longrightarrow sin \theta = 0.5}[/tex]
[tex]\qquad \pmb{\sf\longrightarrow sin \theta = \dfrac{1}{2}}[/tex]
[tex]\qquad \longrightarrow {\pink{\underline{\underline{\pmb{\sf{ \theta = 30^{\circ} }}}}}}[/tex]
Therefore–
- The direction of initial velocity is 30° from the horizontal.
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