From elementary physics and Homework #1, we know that the distance a projectile

Question

From elementary physics and Homework #1, we know that the distance a projectile travels when fired from, eg, a cannon depends on both the initial velocity v and the launch angle theta. x(t) = v t cos(theta) y(t) = v t sin(theta)- 1/2 g t^2 where x(t) and y(t) are the distances traveled in x and y, respectively, after time t has passed (g is gravity, g = 9 8 meters/second") Assume an initial velocity of 13 5 meters per second, and an angle of 89 degrees above the ground Use fzero to find the time and x location where the projectile hits the ground. Print the time and location to 4 decimal places

Explanation / Answer

Given

   the angle of projection is theta = 89 degrees

   the initial velocity u = 13.5 m/s

   acceleration due to gravity is g = 9.8 m/s2

we know that , the horizontal velocity of the projectile never change its constant

the vertical velocity will change ,

the range of the projectile is R = u^2 sin 2theta/g

               R = 13.5^2 sin(2*89)/9.8 m

               R = 0.6490238034725 m

and the time period is

       T = 2 u sin theta /g

       T = 2*13.5sin89/9.8 s

       T = 2.7547 s

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