Question

An object is thrown straight up with a velocity, in ft/s, given by v(t)= -32t + 83, where t is in seconds, from a height of 46 feet.
a) What is the object's initial velocity?
b) What is the object's maximum velocity?
c) What is the object's maximum displacement?
d) When does the maximum displacement occur?
e) When is the object's displacement 0?
f) What is the object's maximum height?

Answer 1

a) Initial velocity = 83 ft/s

b) Object's maximum speed = 99.4 ft/s

c) Object's maximum displacement = 153.64 ft

d) Maximum displacement occur at t = 2.59 seconds.

e) The displacement is zero when t = 5.70 seconds

f) Object's maximum height = 153.64 ft

Explanation:

We have velocity

             v(t)= -32t + 83

Integrating

              s(t) = -16t²+83t+C

At t = 0 displacement is 46 feet

              46 = -16 x 0²+83 x 0+C

                 C = 46 feet

So displacement is

              s(t) = -16t²+83t+46

a) Initial velocity is

                 v(0)= -32 x 0 + 83 = 83 ft/s

       Initial velocity = 83 ft/s

b) Maximum velocity is when the object reaches ground, that is s(t) = 0 ft

Substituting

             0 = -16t²+83t+46

             t = 5.70 seconds

Substituting in velocity equation

           v(t)= -32 x 5.70 + 83 = -99.4 ft/s

           Object's maximum speed = 99.4 ft/s

c) Maximum displacement is when the velocity is zero

   That is

                 -32t + 83 = 0

                       t = 2.59 s

Substituting in displacement equation

                s(2.59) = -16 x 2.59²+83 x 2.59+46 = 153.64 ft

Object's maximum displacement = 153.64 ft

d) Maximum displacement occur at t = 2.59 seconds.

e) Refer part b

   The displacement is zero when t = 5.70 seconds

f) Same as option d

   Object's maximum height = 153.64 ft

Answer 2

Answer:

The initial velocity is 83 ft/s.

The maximum velocity of object is -82.76 ft/s.

The maximum displacement is 107.64 ft.

Time for maximum displacement is 2.59 s.

The object's displacement is zero at 5.18 s.

The maximum height of object is, 107.64 ft.

Explanation:

Given data:

Equation for the velocity is, $v(t)=-32t+83$.

Height is, $H'=46\;\rm feet$.

(a)

At initial, the time function is zero. Which means, t = 0.

Then, initial velocity is:

$v(t=0)=-32(0)+83\\v(t=0)= 83\;\rm ft/s$

Thus, the initial velocity is 83 ft/s.

(b)

The maximum velocity of object is at ground. Then, equation for maximum distance covered is obtained as,

$v(t)=\frac{dH}{dt} \\dH=\int\limits^H_0 {v(t)} \, dt \\dH=\int\limits^H_0 {(-32t+83)} \, dt$

Integrating as,

$H =-16t^2+83t$

Maximum velocity is at ground, hence H=0. Solving as,

$0 =-16t^2+83t\\16t=83\\t=5.18 \;\rm s$

Now, maximum velocity is,

$v(t=5.18)=-32(5.18)+83\\v(t=5.18)=-82.76 \;\rm ft/s$

Thus, maximum velocity of object is -82.76 ft/s.

(c)

The maximum displacement will be at corresponding to zero velocity. Then,

$v(t)=-32t+83\\0=-32t+83\\t=2.59 \;\rm s$

Then, maximum displacement is,

$H =-16(2.59^2)+83(2.59)\\H = 107.64\;\rm ft$

Thus, maximum displacement is 107.64 ft.

(d)

The maximum displacement occurs at zero velocity. And, time is,

$v(t)=-32t+83\\0=-32t+83\\t=2.59 \;\rm s$

Thus, time for maximum displacement is 2.59 s.

(e)

The object displacement is zero when it reaches back to the ground. At ground, H=0. Which means time is,

$0 =-16t^2+83t\\\\16t=83\\t=5.18 \;\rm s$

Thus, object's displacement is zero at 5.18 s.

(f)

The maximum height of object is equal to maximum displacement. Thus, maximum height of object is, 107.64 ft.

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