Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface
-An ideal projectile is launched from the origin at an angle of α radians to the horizontal and an initial speed of 175 ft/sec. Find the position function r(t) for this projectile.
A) $\mathbf { r } ( \mathrm { t } ) = ( 175 \mathrm { t } \sin \alpha ) \mathbf { i } + \left( 175 \mathrm { t } \cos \alpha - 16 \mathrm { t } ^ { 2 } \right) \mathbf { j }$
B) $\mathbf { r } ( \mathrm { t } ) = ( 175 \mathrm { t } \cos \alpha ) \mathbf { i } + \left( 175 \mathrm { t } \sin \alpha - 16 \mathrm { t } ^ { 2 } \right) \mathbf { j }$
C) $\mathbf { r } ( \mathrm { t } ) = \left( 175 \mathrm { t } \sin \alpha - 16 \mathrm { t } ^ { 2 } \right) \mathbf { i } + ( 175 \mathrm { t } \cos \alpha ) \mathbf { j }$
D) $\mathbf { r } ( \mathrm { t } ) = \left( 175 \mathrm { t } \cos \alpha - 32 \mathrm { t } ^ { 2 } \right) \mathbf { i } + ( 175 \mathrm { t } \sin \alpha ) \mathbf { j }$

Question

Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface
-An ideal projectile is launched from the origin at an angle of α radians to the horizontal and an initial speed of 175 ft/sec. Find the position function r(t) for this projectile. 
A) $\mathbf { r } ( \mathrm { t } ) = ( 175 \mathrm { t } \sin \alpha ) \mathbf { i } + \left( 175 \mathrm { t } \cos \alpha - 16 \mathrm { t } ^ { 2 } \right) \mathbf { j }$
B) $\mathbf { r } ( \mathrm { t } ) = ( 175 \mathrm { t } \cos \alpha ) \mathbf { i } + \left( 175 \mathrm { t } \sin \alpha - 16 \mathrm { t } ^ { 2 } \right) \mathbf { j }$
C) $\mathbf { r } ( \mathrm { t } ) = \left( 175 \mathrm { t } \sin \alpha - 16 \mathrm { t } ^ { 2 } \right) \mathbf { i } + ( 175 \mathrm { t } \cos \alpha ) \mathbf { j }$
D) $\mathbf { r } ( \mathrm { t } ) = \left( 175 \mathrm { t } \cos \alpha - 32 \mathrm { t } ^ { 2 } \right) \mathbf { i } + ( 175 \mathrm { t } \sin \alpha ) \mathbf { j }$

Quizplus · Accepted Answer

We can use the following kinematic equations to find the position function r(t):
r(t) = r0 + v0t + (1/2)at^2

Since the projectile is launched from the origin, we have r0 = (0,0).

The initial velocity vector is given by:
v0 = v0x i + v0y j
where v0x = v0 cos(α) and v0y = v0 sin(α).

Thus, we have v0 = 175 cos(α) i + 175 sin(α) j.

The acceleration due to gravity is a = -32.2 ft/sec^2 in the vertical direction (assuming we are working in imperial units).

Plugging in these values in the kinematic equation for the y-component (vertical direction), we get:
y(t) = v0y t + (1/2)at^2
      = 175 sin(α) t - 16t^2

Plugging in the values in the kinematic equation for the x-component (horizontal direction), we get:
x(t) = v0x t
      = 175 cos(α) t

Thus, the position function r(t) is:
r(t) = x(t) i + y(t) j
      = (175 t cos(α)) i + (175 t sin(α) - 16t^2) j

Therefore, the answer is B.

Solve the Problem B) C) D)