A stone is catapulted at time t = 0, with an initial velocity of magnitude 20.2 m/s and at an angle of 39.1° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.10 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.73 s.
Call the horizontal displacement X and the vertical displacement Y.
X = 20.2 cos39.1 * t
Y = 20.2 sin39.1 * t - 4.9 t^2
Plug in the appropriate times (t) to get the answers they want.
To find the magnitudes of the horizontal and vertical components of displacement at different times, we can use the equations of projectile motion. Let's break down the problem into parts:
Given:
Initial velocity magnitude (V₀) = 20.2 m/s
Launch angle (θ) = 39.1°
(a) To find the horizontal component of displacement at t = 1.10 s:
We can use the formula:
Horizontal component (X) = V₀ * cos(θ) * t
Plugging in the values:
X = 20.2 m/s * cos(39.1°) * 1.10 s
Now we can calculate X.
(b) To find the vertical component of displacement at t = 1.10 s:
We can use the formula:
Vertical component (Y) = V₀ * sin(θ) * t - 0.5 * g * t²
Plugging in the values:
Y = 20.2 m/s * sin(39.1°) * 1.10 s - 0.5 * 9.8 m/s² * (1.10 s)²
Now we can calculate Y.
Repeat the above steps to find the horizontal and vertical components of displacement at t = 1.73 s:
(c) Horizontal component:
X = 20.2 m/s * cos(39.1°) * 1.73 s
(d) Vertical component:
Y = 20.2 m/s * sin(39.1°) * 1.73 s - 0.5 * 9.8 m/s² * (1.73 s)²
Now you can calculate X and Y for both cases.