Astone is catapulted at time t = 0, with an initial velocity of magnitude 19.6 m/s and at an angle of 40.8° above the horizontal. what are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.03 s? repeat for the (c) horizontal and (d) vertical components at t = 1.73 s, and for the (e) horizontal and (f) vertical components at t = 5.05 s. assume that the catapult is positioned on a plain horizontal ground.

To answer this question, let us break down the initial velocity into its horizontal and vertical components.

v_{ix} = vcos(θ)

v_{iy} = vsin(θ)

v_{x} is the horizontal component of the initial velocity, and this value will stay constant over time because we assume only gravity acts on the stone, therefore only the vertical component of the stone's velocity will change over time

v_{iy} is the vertical component of the initial velocity, and this will change over time due to gravity.

v is the magnitude of the initial velocity.

θ is the angle the velocity vector is oriented at with respect to the horizontal.

Given values:

v = 19.6m/s

θ = 40.8°

Plug in these values and solve for v_{ix} and v_{iy}:

v_{ix} = 19.6cos(40.8°) = 14.8m/s

v_{iy} = 19.6sin(40.8°) = 12.8m/s

To find both the horizontal and vertical displacement at any time, we will use this kinematics equation:

D = X + Vt + 0.5At²

When solving for the horizontal displacement, the following values are:

t = elapsed time

D = horizontal displacement

X = initial horizontal displacement

V = initial horizontal velocity

A = horizontal acceleration

There is no initial horizontal displacement, so X = 0m

The initial horizontal velocity V = v_{ix} = 14.8m/s

Assuming we don't care about air resistance, no force has a component acting horizontally on the stone, so A = 0m/s²

Therefore the equation for the stone's horizontal displacement is given by:

D = 14.8t

When solving for the vertical displacement, the following values are:

t = elapsed time

D = vertical displacement

X = initial vertical displacement

V = initial vertical velocity

A = vertical acceleration

There is no initial vertical displacement, so X = 0m

The initial vertical velocity V = v_{iy} = 12.8m/s

Gravity acts downward on the stone, therefore A = -9.81m/s²

Therefore the equation for the stone's vertical displacement is given by:

D = 12.8t - 4.905t²

Now we just plug in various values of t...

a) At t = 1.03s, the horizontal displacement is D = 14.8(1.03) = 15.2m

b) At t = 1.03s, the vertical displacement is D = 12.8(1.03)-4.905(1.03)² = 7.98m

c) At t = 1.73s, the horizontal displacement is D = 14.8(1.73) = 25.6m

d) At t = 1.73s, the vertical displacement is D = 12.8(1.73)-4.905(1.73)² = 7.46m

Before you write down the following results, read the following explanation.

e) At t = 5.05s, the horizontal displacement is D = 14.8(5.05) = 74.7m

f) At t = 5.05s, the vertical displacement is D = 12.8(5.05)-4.905(5.05)² = -60.4m

Notice that we have a negative value for the vertical displacement. This isn't possible within the context of the problem, so the vertical displacement at t = 5.05s is actually 0m

The problem hasn't stated whether the ground is frictionless or not. I'm going to assume it is frictionless and therefore the stone will keep moving across the ground after landing, so the horizontal displacement at t = 5.05s is 74.7m