A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is 60 i + 64 k, with speed measured in feet per second. The spin of the ball results in a southward acceleration of 6 ft/s2, so the acceleration vector is a = −6 j − 32 k. Where does the ball land? (Round your answers to one decimal place.) ft from the origin at an angle of ° from the eastern direction toward the south. With what speed does the ball hit the ground? (Round your answer to one decimal place.)

kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is

        r = 244.8 ft, tea = 21.8º from East to South

        v = 91.0 ft / s

given parameters

to find

Kinematics allows finding the position, velocity and acceleration of the body, in this case we have a problem in three dimensions, where they establish a Cartesian coordinate system, a method to solve this exercise is to solve each component independently

a)  The acceleration of gravity acts on the z axis, so let's find the time it takes to reach the ground, if the initial vertical velocity is v_{oz} = 64 ft/s and the acceleration is a_z = g = -32 ft / s², we assume that the ball leaves the ground (z₀ = 0)

         z = z₀ + v_{oz} t + ½ a_z t²

when reaching the ground its height of zero and

        0 = 0 + v_{oz} t + ½ a_z t²

        t (v_{oz) + ½ a_z t) = 0

        t (64 - 16 t) = 0

the solusion of this squadron is

       y = 0

       t = 4 s

the first time is when it leaves and the second time is for when it reaches the ground, therefore the flight time is t = 4s

with this time we find the displacement is each exercise

X axis

    in this axis there is no acceleration, so we use the uniform motion relationships

        vₓ = x / t

        x = vₓ t

        x = 60 4

        x = 240 ft

Y Axis

on this axis there is an acceleration of a_y = -6 ft/s and an initial velocity v_{oy} = 0

we use the kinematic relation

       y = v_{oy} t + ½ a_y t²

       y = 0 - ½ 6 4²

       y = - 48 ft

let's use the Pythagoras theorem to find the position

       r² = (x -x₀) ² + (y -y₀) ² + (z -z₀) ²

       r² = (240-0) ² + (-48-0) ² + (0-0) ²

       r = 244.8 ft

We use trigonometry to find the direction

      tan θ = y / x

       θ = tan⁻¹

       θ = tan⁻¹  

       θ = -21.8º

This angle is measured clockwise from the x axis, it can also be read

        θ = 21.8º from East to South

b)  Let's look for the speed when we hit the ground

X axis

       vₓ = v_{ox} + aₓ t

       vₓ = 60 - 0

       vₓ = 60 ft / s

Y Axis

       v_y = v_{oy} + a_y t

       v_y = 0 - 6 4

        v_y = -24 ft / s²

Z axis

       v_z = v_{oz} + a_z t

       v_z = 64 -32 4

       v_z = -64 ft / s

With the Pytagoras  theorem find the modulus of this speed is

        v² = vx² + vy² + vz²

        v² = 60² + 24 ² + 64²

        v = 91.0 ft / s

In conclusion, the kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is

       a) r = 244.8 ft, θ = 21.8º from East to South

       b) v = 91.0 ft / s

learn more about kinematics here:

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The ball's position in the air at time t is given by the vector,

p(t) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ²

and its velocity is given by

v(t) = (60 i + 64 k) + (-6 j - 32 k) t

The ball is in the air for as long as it takes for the vertical (k) component of the position vector to reach 0, so we solve,

64 t - 32/2 t ² = 0  ==>  t = 0 OR t = 4

and so the ball is in the air for 4 s.

After this time, the ball has position vector

p(4) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ² = 240 i - 48 j

which has magnitude

||p(4)|| = √(240² + (-48)²) = 48 √26 ≈ 244.8 ft

in a direction θ in the x,y plane from the positive x axis such that

tanθ = -48/240 = -1/5  ==>  θ = -arctan(1/5) ≈ -11.3º

or 11.3º South of East.

The ball hits the ground with speed

||v(4)|| = ||60 i - 24 j - 64 k|| = √(60² + (-24)² + (-64)²) = 4 √517 ≈ 91.0 ft/s

gans

explanation:

vestigial organs (like your appendix) are organs that no longer serve a purpose.

b) phosphorus pentoxide