Question

A ball is thrown vertically upward from a height of 5 ft with an initial velocity of 83 ft/s. The distance s (in feet) of the ball from the ground after t seconds is s = s(t) = 5 + 83t − 16t^2. (a) What is the velocity v, in feet per second, of the ball after 2 seconds?

(b) When, in seconds, will the ball reach its maximum height? (Round your answer to one decimal place.)

(c) What is the maximum height, in feet, the ball reaches? (Round your answer to one decimal place.)

(d) What is the acceleration a, in feet per second per second, of the ball at any time t? a(t) = ft/s2 (e) How long, in seconds, is the ball in the air? (Round your answer to three decimal places.) t = s

(f) What is the velocity of the ball, in feet per second, upon impact with the ground? (Round your answer to two decimal places.) v = ft/s What is its speed in feet per second? (Round your answer to two decimal places.)

(g) What is the total distance traveled, in feet, by the ball? (Round your answer to one decimal place.) ft

Answer #1

A
ball is thrown directly upward from a height of 3 ft with an
initial velocity of 20 ft/sec. the function s(t) = -16t^2+20t+3
gives the height of the ball, in feet, t seconds after it has been
thrown. Determine the time at which the ball reaches its maximum
height and find the maximum height.

if a ball is thrown vertically upward with a velocity of 95
ft/s, then its height after t seconds is
s=95t-19t^2
A. what is maximum height reached by the ball?
B. what is the velocity of the ball if its 114 ft above the
ground on the way up?
C. what is the velocity of the ball when it is 114 above the
ground on its way down?

A ball is thrown across a playing field from a height of h = 5
ft above the ground at an angle of 45° to the horizontal at the
speed of 20 ft/s. It can be deduced from physical principles that
the path of the ball is modeled by the function
y= -(32/400)x^2+x+5
where x is the distance in feet that the ball has traveled
horizontally.
1) Find the maximum height attained by the ball. (Round your answer
to three...

If a ball is thrown upward with a velocity of 64 ft/sec, then
its height after t seconds is s=64t-16t^2. When does the ball reach
its maximum height?

A
ball is thrown straight up from the too of a building 128 ft. tall
with qn initial velocity of 48 ft per second. The distance s(t) (in
feet) of the ball from the fround is given by s(t)=128+48t-16t^2.
Find the maximum height attained by the ball.

if a baseball is thrown vertically upward with an initial
velocity of 128 ft/sec, the ball’s height after t seconds is
s(t)=128t-16t^2
A. Find the baseball's velocity and acceleration at time
t.
B. How long does it take the baseball to reach its highest
point?
C. How high does the baseball go?
D. What is the velocity of the ball when it is 112 ft above
the ground on the wh down?
E. What is the speed of the ball...

11. a ball is thrown up from the ground with an initial velocity
of 120 feet per second. The height s in feet can be expressed as a
function of time t in seconds by s(t)=85t-16t^2 a. when does the
ball reach its maximum height b. what is the maximum height c. when
does the ball hit the ground
12. a farmer has 2800 feet of fencing material with which to
enclose a rectangle field that borders a straight stream....

17) A ball is thrown vertically upward from ground level with an
initial velocity of 96 feet per second. Use ?(?) = −32 ??⁄ 2.
a) How long will it take it to rise to its maximum height? b)
What is the maximum height?

Use the position function
s(t) =
−16t2 +
v0t +
s0
for free-falling objects.
A ball is thrown straight down from the top of a 500-foot
building with an initial velocity of −40 feet per second.
(a)
Determine the position and velocity
v(t) functions for the
ball.
s(t)
=
v(t)
=
(b)
Determine the average velocity, in feet per second, on the
interval
[1, 2].
ft/s
(c)
Find the instantaneous velocities, in feet per second, when
t = 1and t...

If an object is propelled upward from a height of s feet at an
initial velocity of v feet per second, then its height h after t
seconds is given by the equation h=−16t^2+vt+s, where h is in feet.
If the object is propelled from a height of 1212 feet with an
initial velocity of 9696 feet per second, its height h is given by
the equation h =minus−16t^2+96 +12. After how many seconds is the
height 120 feet?

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