Question

Basic Physics PHY 114 -W-01 Department of Physical and Computer Sciences Medgar Evers College Workshop Lab 5: Projectiles Name: Name: Name: lue triangle), The diagram below illustrates the scene of a ship (green triagle), a mountain (b and a target (red triangle). CR For each problem, you will be given the values of ro,ER, and H along with two angles 0 and 92 The angles indicates the orientation of ship G's cannon when fired. You must find 1. The maximum height reached by the projectile fired at each angle. 2 The possible maximum distance traveled by the projectile fired at each angle. 3. State whether the projectile did not scale (NS) the mountain (blue), was short (S) of the target (red), was over (O) the target, or hit (Y) the target. 4. Last (final assessment), if the target was missed, give an angle that can reach the target or determine that the target cannot be reached with the current values. The speed of the cannon's projectile is always 250 m/s.

Answer 1

theta_1 = 30 degree u = 250 m/s Total distance = 3000 m Maximum height (H) = u^2 sin^2 theta/2g = 250^2 sin^2 30/2 times 9.8 m H = 797.19 m Maximum distance (R) = u^2 sin 2 theta/g = 250^2 sin 2 times 30/9.8 Therefore R = 250^2 sin 60/9.8 = 5523.12 m Projectile did not scale the mountain. Projectile would have been the target. From y = (tan theta)x - 1/2 times g/u^2 cos^2 theta x^2 = tan 30 times 2000 - 1/2 times 9.8/(250 cos 30)^2 times (2000)^2 y = 1154.7 - 418.13 = 736.57 m \r\n For theta_2 = 60 degree Maximum height (H) = u^2 sin^2 theta/2g Therefore H = 250^2 sin^2 60/2 times 9.8 m = 2391.58 m Maximum distance (R) = u^2 sin 2 theta/g R = 250^2 sin 2 times 60/9.8 m = 5523.12 m Therefore Projectile did not the mountain Projectile would have been over the tangent.