A projectile is fired at time t = 0.0 s,from point 0 at the edge of a cliff,with initial velocity components of v_ox = 50 m/s and v_oy =200 m/s.The projectile rises,then falls into the sea at point P.The time of flight of the projectile is 50.0 s. In the figure,the y-coordinate of the projectile when its x-coordinate is 1000 m is closest to: A)+ 2040 m B)150 m C)- 150 m D)+ 1240 m E)+ 600 m

Question

Quizplus · Accepted Answer

We can use the equations of motion to find the position of the projectile at any time t. Let's choose the origin to be at point 0, and take the positive y-direction to be upwards. Then we have: y = v_oy t - 0.5 g t^2 x = v_ox t where g is the acceleration due to gravity, and we have neglected air resistance. Since the time of flight is 50.0 s, the projectile will hit the sea when y = 0 at some value of x that we can find. Substituting y = 0 into the equation for y, we get: 0 = v_oy t - 0.5 g t^2 t = 2 v_oy / g Substituting this value of t into the equation for x, we get: x = v_ox (2 v_oy / g) x = (v_ox v_oy / g) x = (50 m/s)(200 m/s) / (9.81 m/s^2) x = 1019.2 m Therefore, the x-coordinate when the projectile hits the sea is approximately 1019.2 m. To find the y-coordinate at x = 1000 m, we can use the equation for y: y = v_oy t - 0.5 g t^2 We need to solve for t in terms of x, and then substitute x = 1000 m to find y. Using the equation for x above, we can solve for t: t = x / (v_ox) Substituting x = 1000 m and simplifying, we get: t = 20 s Substituting t = 20 s into the equation for y, we get: y = (200 m/s)(20 s) - 0.5 (9.81 m/s^2) (20 s)^2 y = 4000 m - 1962 m y = 2038 m Therefore, the y-coordinate of the projectile when its x-coordinate is 1000 m is closest to 2040 m (choice A).

A Projectile Is Fired at Time T = 0