1. A small dense ball with mass 1.5 kg is thrown with an initial velocity (5, 8,

Question

1. A small dense ball with mass 1.5 kg is thrown with an initial velocity (5, 8, 0) m/s, starting at the origin. For this problem neglect air drag.

a. What is the velocity vector when the height is a maximum (at the top)?

b. At what time does it reach the top?

c. What is the average velocity during this time interval?

d. What is the position (vector) at the top?

e. At what time does the ball’s height return to zero?

f. What is the x- position when the height returns to zero (aka the range)

g. What is vy at this time (in part (e))?

h. What was the angle to the x-axis of the initial velocity?

i. What was the angle to the x-axis of the velocity at the time in part (e)?

Explanation / Answer

mass = m = 1.5 kg

vox = 5

voy = 8

voz = 0

(a)

at the maximum height vertical component become zero lezving horizantal component

vx = ( 5 0 0 )

(b)

from equation of motion

vy = voy + gt

0 = 8 -10*t

t = 0.8

t = 0.8 s

(c)

Vavg = Vi + Vf /2

Vavg = ((5 , 8, 0) + (5 ,0 ,0 ) ) /2 = (5 , 4 , 0 )

(d)

ri = vi*t + 0.5*ay*t^2

ri = (5i + 8j )*0.8 - 0.5*10*0.8^2*j

ri = 4.i - 3.2 j

(e)

by the time the ball returns to zero . y = 0

y = voy*T + 0.5*g*T^2

0 = 8*T - 0.5*10*T^2

T = 1.6 s

(f)

x = vox *T= 5*1. 6 = 8 m

g)

vy = voy + gT = 8 - 10*1.6 = -8 m/s

h)

theta = tan^-6(voy/vox) = tan^-1(8/5) = 58

i)

theta = tan^-1(vy/vx) = -58

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