A spring-loaded gun can fire a projectile to a height h if it is fired straight up. If the same gun is pointed at an angle of 45° from the v

Question

A spring-loaded gun can fire a projectile to a height h if it is fired straight up. If the same gun is pointed at an angle of 45° from the vertical, what maximum height can now be reached by the projectile? h4

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MichaelMet 2 months 2021-07-28T08:33:49+00:00 1 Answers 5 views 0

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    2021-07-28T08:35:29+00:00

    Answer:

    The maximum height can  be reached by the projectile h_{2} = \frac{h}{2}

    Explanation:

    Height = h

    \theta = 45

    Height in a projectile motion is given by

    h = \frac{u^{2} }{2 g} \sin^{2}  \theta

    When gun fired  straight up then \theta = 90°

    In that case Height

    h = \frac{u^{2} }{2 g} \sin^{2}  90

    h = \frac{u^{2} }{2 g} —— (1)

    When \theta = 45° then

    h_2 = \frac{u^{2} }{2 g} \sin^{2}  45

    h_2 = \frac{u^{2} }{2 g} (\frac{1}{2} )

    From equation (1)

    h_{2} = \frac{h}{2}

    This is the maximum height can  be reached by the projectile.

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