a projectile is fired at an angle of 60 with the gound with a speed of 100m/s.the time taken for the inclination with the ground becomes to be 45 is
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Imagine the projectile being thrown from the origin of a graph
When the inclination is 45° the x coordinate and y coordinate will be equal
If the initial velocity of the projectile is taken as u, then we know
[tex]x=(ucos\theta)t \\ y=(usin\theta)t-\frac{1}{2}gt^{2}[/tex]
When x=y
[tex](ucos\theta)t = (usin\theta)t-\frac{1}{2}gt^{2} \\ (ucos\theta) = usin\theta-\frac{1}{2}gt \\ \\ t = \frac{2u(sin\theta-cos\theta)}{g} \\ \\ In the given case u = 100 m/s and \theta = 60^{0} \\ [/tex]
Substituting the values
If we take the value of g as 10 m/s²
When the inclination is 45° the x coordinate and y coordinate will be equal
If the initial velocity of the projectile is taken as u, then we know
[tex]x=(ucos\theta)t \\ y=(usin\theta)t-\frac{1}{2}gt^{2}[/tex]
When x=y
[tex](ucos\theta)t = (usin\theta)t-\frac{1}{2}gt^{2} \\ (ucos\theta) = usin\theta-\frac{1}{2}gt \\ \\ t = \frac{2u(sin\theta-cos\theta)}{g} \\ \\ In the given case u = 100 m/s and \theta = 60^{0} \\ [/tex]
Substituting the values
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