why do yo have to subtract 53 from 90 to get a negative37degrees to solve this p

Question

why do yo have to subtract 53 from 90 to get a negative37degrees to solve this problem? A plane diving with constant speed at an angle of 53 degreeswith the verticle, releases a projectile at an altitude of 730m.The projectile hits the ground 5.0s after release. What is thespeed of the plane? why do yo have to subtract 53 from 90 to get a negative37degrees to solve this problem? A plane diving with constant speed at an angle of 53 degreeswith the verticle, releases a projectile at an altitude of 730m.The projectile hits the ground 5.0s after release. What is thespeed of the plane?

Explanation / Answer

You don't have to subtract 53 -90 = -37 degrees . Use the vertical axis as your reference h.i = 730 m v.y = - v cos 53 v.x = v sin 53 g = 9.81 m . The equation of motion in the vertical direction is h = h.i + v.y t - 1/2 g t^2 Set h = 0 for when the bomb hits 0 = h.i + v.y t - 1/2 g t^2 0 = 730m + (-v cos(53)) * 7 sec - 1/2 * 9.81 m/s^2 * 49sec^2 v cos(53) * 7 sec = 730m - 240.345 m v = (489.66m)/(cos53 * 7sec) v = 116.23 m/s

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