Question

A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is 40 i + 32 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.) ft/s

Answer 1

The answer is "52.61 ft/s"

Step-by-step explanation:

Taking the motion in the z-direction is
`u_z = 32 , \ \ a_z = - 32`

`t = 2 (u_z)/(( - a_z ))\\\\`

`= 2 * (32)/(32)\\\\ = 2 \ sec\\\\`

z-axis change is zero z and the x-path the same way:

`= u_x t + (1)/(2) a_x t^2 \\\\= 40 * 2 + 0.5 * 0 * 2^2 \\\\ = 40 * 2 + 0.5 * 0 * 4 \\\\ = 80 +0\\\\= 80 \ ft \ \ i`

In y-direction:

`= 0 * t - (1)/(2) * (6) * 2^2\\\\ = - (1)/(2) * (6) * 4\\\\ = - (6) * 2\\\\= - 12 ft \ \ j`

the ball lands at:
`( 80 i - 12 j + 0 k ) ft`

Magnitude:

`\to ( 80^2 + (-12)^2 )^{(1)/(2)}\\\\\to ( 6400 + 144 )^{(1)/(2)}\\\\\to ( 6544 )^{(1)/(2)}\\\\\to 80.89 \ ft`from the origin.

angle:

`\to \tan^(-1) ( (12)/(80) )\\\\\to 8.53^(\circ)`

velocity:

`\to 40 i + ( 0 - 6 * 2 ) j + ( 32- 32 * 2 ) k\\\\\to 40 i + ( - 12 ) j + ( 32- 64 ) k\\\\\to 40 i - 12 j - 32 k\\\\\to ( 40^2 + 12^2 + 32^2 )^{(1)/(2)}\\\\\to (1600+144+1024)^{(1)/(2)}\\\\\to 2768^{(1)/(2)}\\\\\to 52.61\ \ (ft)/(s)`