Initially (at time t = 0) a particle is moving vertically at 5.8 m/s and horizontally at 0 m/s. Its horizontal acceleration is 2.1 m/s^2. At what time will the particle be traveling at 54◦ with respect to the horizontal? The acceleration due to gravity is 9.8 m/s^2
To solve this problem, we need to break down the initial velocity into its horizontal and vertical components.
Given:
Initial horizontal velocity (u_x) = 0 m/s
Horizontal acceleration (a_x) = 2.1 m/s^2
Initial vertical velocity (u_y) = 5.8 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
First, let's calculate the time it will take for the particle to reach the angle of 54° with respect to the horizontal.
Step 1: Find the vertical velocity component at 54°
Given angle (θ) = 54°
Vertical velocity component (v_y) = u_y - g * t
v_y = 5.8 m/s - 9.8 m/s^2 * t
Step 2: Find the horizontal velocity component at 54°
Horizontal velocity component (v_x) = u_x + a_x * t
Since the initial horizontal velocity (u_x) is 0 m/s, the horizontal velocity component remains 0 m/s throughout.
Step 3: Calculate the total velocity at 54°
The total velocity (v_total) at 54° is given by the Pythagorean theorem:
v_total^2 = v_x^2 + v_y^2
Substituting the values:
v_total^2 = 0^2 + (5.8 - 9.8t)^2
Step 4: Calculate the time taken to reach the angle of 54°
Since we need to find the time when the particle will be traveling at 54°, we can set v_total equal to the magnitude of the initial velocity (5.8 m/s) and solve for t.
5.8^2 = (5.8 - 9.8t)^2
Simplifying the equation:
33.64 = 33.64 - 114.64t + 96.04t^2
Rearranging the equation:
96.04t^2 - 114.64t = 0
Dividing through by 2.56:
37.51t^2 - 44.92t = 0
Factoring out t:
t(37.51t - 44.92) = 0
Either t = 0 (which is the initial time) or:
37.51t - 44.92 = 0
37.51t = 44.92
t = 1.2 seconds (to two significant figures)
Therefore, the particle will be traveling at 54° with respect to the horizontal after approximately 1.2 seconds.
To find the time at which the particle will be traveling at 54 degrees with respect to the horizontal, we need to first break down the particle's initial velocity into its vertical and horizontal components.
Given:
Initial vertical velocity (Vy) = 5.8 m/s
Initial horizontal velocity (Vx) = 0 m/s
Horizontal acceleration (Ax) = 2.1 m/s^2
Acceleration due to gravity (g) = 9.8 m/s^2
We know that the vertical and horizontal motions of the particle are independent of each other. So, let's start with the horizontal motion.
Since the particle has an initial horizontal velocity of 0 m/s and a constant horizontal acceleration of 2.1 m/s^2, we can use the kinematic equation:
Vx = Vxi + Ax*t
Since Vxi = 0 m/s (initial horizontal velocity), the equation simplifies to:
Vx = Ax*t
Now, let's move on to the vertical motion. The particle starts with an initial vertical velocity (Vy) of 5.8 m/s and experiences a constant acceleration due to gravity (g) of -9.8 m/s^2 (negative sign because the acceleration is directed opposite to the initial velocity).
Using the kinematic equation for the vertical motion:
Vy = Vyi + g*t
Since Vyi = 5.8 m/s (initial vertical velocity), the equation becomes:
Vy = 5.8 m/s - 9.8 m/s^2 * t
Next, we need to find the combined velocity (V) of the particle at any given time (t). The combined velocity can be calculated using the Pythagorean theorem:
V = sqrt(Vx^2 + Vy^2)
Since we want to know the time at which the particle is traveling at 54 degrees with respect to the horizontal, we can set up an equation:
tan(54) = Vy / Vx
tan(54) = [5.8 - 9.8 * t] / (2.1 * t)
Now, we can solve this equation to find the value of t.
tan54 = Y/X,
tan54 = 5.8 / X,
X = 5.8 / tan54 = 4.21m/s.
t = (Vf - Vo) / a,
t = (4.21 - 0) / 2.1 = 2.0s.