Initially (at time t = 0) a particle is moving
vertically at 7.2 m/s and horizontally at
0 m/s. Its horizontal acceleration is 1.3 m/s
2
.
At what time will the particle be traveling
at 59◦ with respect to the horizontal? The
acceleration due to gravity is 9.8 m/s
2
.
Answer in units of s.
019 (part 1 of 2) 10.0 po
To find the time at which the particle will be traveling at 59° with respect to the horizontal, we can break the particle's motion into horizontal and vertical components.
Given:
Initial vertical velocity (Viy) = 7.2 m/s
Initial horizontal velocity (Vix) = 0 m/s
Horizontal acceleration (Ax) = 1.3 m/s^2
Acceleration due to gravity (Ay) = 9.8 m/s^2
Since there is no horizontal acceleration (Ax = 0), the horizontal velocity (Vx) remains constant throughout the motion.
The vertical motion of the particle can be analyzed using the equations of motion under constant acceleration:
Vertical displacement (Δy) = Viy*t + (1/2)Ay*t^2
Vertical velocity (Viy) = Viy + Ay*t
At the instant when the particle is traveling at 59° with respect to the horizontal, the vertical and horizontal velocities can be related using trigonometry:
tan(59°) = Viy / Vx
Substituting the known values:
tan(59°) = 7.2 / Vx
Simplifying, we find:
Vx = 7.2 / tan(59°)
Now, since Vx is constant, we can equate Vx to the horizontal velocity Vix = 0:
Vix = Vx
Therefore, solving for Vix:
0 = 7.2 / tan(59°)
Simplifying,
tan(59°) = 7.2 / 0
This is not a valid expression as we cannot divide by zero. Hence, it shows that the particle will never have a horizontal velocity of zero while moving at 59° with respect to the horizontal.
Therefore, there is no time at which the particle will be traveling at 59° with respect to the horizontal.