Question

At time t=0, a ball is thrown straight down from the top of a 384-foot building with an initial velocity of -32 feet/second.

(a) When does the ball hit the ground?
(b) What is the ball's velocity after falling 128 feet?
(c) What is the ball's velocity at impact?

Note: This problem requires the equation:
s(t) = -16t^2 + v₀t + s₀

Answer 1

first of all, t=0 s=38ft (down), a=9.8m/s^2 (down), initial vel = -32ft/s and final vel = ?

Answer 2

we need to convert the units to be the same, 1ft=0.3048m, therefore s=11.6m, intial vel = -9.75m/s

Answer 3

It is common to use 32 ft/sec^2 for the acceleration of gravity

Answer 4

@Lion when you write s(t) do you mean ds/dt ?

Answer 5

s(t) means distance. In (very) old text books s was short for "space"

Answer 6

so we get a quadratic equation for the t=0,
\[-16t ^{2}-32t+384=0\]

Answer 7

t=-6 or 4

Answer 8

therefore the ball hits the ground after 4s