A projectile is projected from the origin with a velocity of 30.0 m/s at an angle of 40.0 degrees above the horizontal. What is the position vector when the projectile is at the maximum height?
A) X = 45.2 m, Y = 19.0 m
B) X = 21.1 m, Y = 19.0 m
C) X = 51.1 m, Y = 21.1 m
D) X = 45.2 m, Y = 21.1 m
E) X = 19.0 m, Y = 17.5 m

Question

Quizplus · Accepted Answer

The correct answer is A because the projectile reaches its maximum height when its vertical velocity component is zero. At this point, the projectile has traveled a horizontal distance of 30.0 m/s * cos(40.0 degrees) * t, where t is the time it takes to reach the maximum height. The vertical component of the projectile's velocity is 30.0 m/s * sin(40.0 degrees), and the acceleration due to gravity is -9.8 m/s^2. Using the equation v = u + at, we can find that the time it takes to reach the maximum height is t = 30.0 m/s * sin(40.0 degrees) / 9.8 m/s^2 = 1.90 s. Therefore, the horizontal distance traveled by the projectile is 30.0 m/s * cos(40.0 degrees) * 1.90 s = 45.2 m. The maximum height reached by the projectile is 30.0 m/s * sin(40.0 degrees) * 1.90 s - 0.5 * 9.8 m/s^2 * (1.90 s)^2 = 19.0 m.

A Projectile Is Projected from the Origin with a Velocity