Respuesta :
Answer:
The maximum height can be reached by the projectile [tex]h_{2} = \frac{h}{2}[/tex]
Explanation:
Height = h
[tex]\theta = 45[/tex]
Height in a projectile motion is given by
[tex]h = \frac{u^{2} }{2 g} \sin^{2} \theta[/tex]
When gun fired straight up then [tex]\theta[/tex] = 90°
In that case Height
[tex]h = \frac{u^{2} }{2 g} \sin^{2} 90[/tex]
[tex]h = \frac{u^{2} }{2 g}[/tex] ------ (1)
When [tex]\theta = 45[/tex]° then
[tex]h_2 = \frac{u^{2} }{2 g} \sin^{2} 45[/tex]
[tex]h_2 = \frac{u^{2} }{2 g} (\frac{1}{2} )[/tex]
From equation (1)
[tex]h_{2} = \frac{h}{2}[/tex]
This is the maximum height can be reached by the projectile.
