A ball at rest begins to roll down a hill with a 12-degree slope for a total of 9 seconds. A) How far did the ball move from the origin after 4 seconds? B) What is the ball's velocity after 5 seconds?

Question

A ball at rest begins to roll down a hill with a 12-degree slope for a total of 9 seconds.

A) How far did the ball move from the origin after 4 seconds?

B) What is the ball's velocity after 5 seconds?

Answer 1

12 degree slope

how far in 4 seconds?

velocity in 5 seconds?

gravity has a force of negative 32 ft/sec/sec at sea level

vertical height =h(t)= -16t^2

in 9 seconds, h= h(9) = -16(9^2) = -16(81) = - 1296 feet

convert that to slope distance by constructing a right triangle

where 1296 =hypotenuse=h, opposite side = distance along the slope =d

= 1296(d/h)=1296sin12 = about 269.45 feet in 9 seconds

if slope had been zero degrees, it wouldn't have moved at all, as 1296sin0=1296(0)=0 feet

a) in 4 seconds, h=-16(4^2) =-256

d= 256sin12 =about 53.23 feet

b) in 5 seconds, velocity downward = -32 ft/sec

d' = velocity along the slope = 32sin12 = about 6.65 feet per second after 5 seconds

assuming the hill slope is a frictionless plane, no wind or obstructions, and a perfectly spherical ball

but someone else came up with

a) 16.3 and b) 10.2

so ?? wait? did they just use meters instead of feet? at least for a) part?

this problem is a problem

and this just in, another guy came up with

11.7 meters and 7.28 meters per second

convert that to feet and ft/sec and what do you get?

wait, last minute update, last guy says he made a mistake

while most though, even those with doctorates in math or engineering seem to pass on even trying this problem. why?

another guy answered d=80,v =20 for 4 seconds, and for 5 seconds d=125 meters and v= 25 meters per second, using a=10 m/s^2 which is a rough approximation for gravity's acceleration