initial velocity of ball is 60 ft/sec upward

[abby_07]

can someone please help me with this problem i do not know if i am approaching it the right way.

use a(t)=-32 ft/sec^2 as the acceleration due to gravity. (Neglect air risistance)

The question states a ball is thrown vertically upward from the ground with an initial velocity of 60 feet per second. How high will the ball go?

what i did was use the equation f(t)=-16t^2+Vot+So

i got Vot to equal 60
and So to equal 0

i got the derivative of f(t)
and got

f'(t)=-32t+60+0

is this equation right? and if it is do i just solve for t?

thank you

 

[TchrWill]

Re: initial velocity

can someone please help me with this problem i do not know if i am approaching it the right way.

use a(t)=-32 ft/sec^2 as the acceleration due to gravity. (Neglect air risistance)

The question states a ball is thrown vertically upward from the ground with an initial velocity of 60 feet per second. How high will the ball go?

what i did was use the equation f(t)=-16t^2+Vot+So

i got Vot to equal 60
and So to equal 0

i got the derivative of f(t)
and got

f'(t)=-32t+60+0

The time to zero velocity derivrs from Vf = Vo - gt where Vf = 0, Vo = 60 and g = 32 making t = 1.875 sec.

The height reached derives from h = Vot - 16t^2 or h = 60(1.875) - 16(1.875)^2 = 56.25 ft.

is this equation right? and if it is do i just solve for t?

thank you

 

[soroban]

Re: initial velocity

Hello, Abby!

use \(\displaystyle a(t)\,=\,-32\) ft/sec² as the acceleration due to gravity. (Neglect air resistance)

A ball is thrown vertically upward from the ground with an initial velocity of 60 ft/sec.
How high will the ball go?

Since they started you with the acceleration,
. . I suspect you are expect to derive the velocity and height functions.

We are given: \(\displaystyle \:a(t) \:=\:-32\)

Integrate: \(\displaystyle \:v(t) \;= \;\int (-32)dt \:=\:-32t\,+\,C_1\;\) [1]

We are told that the initial velocity is 60 ft/sec.
. . That is: \(\displaystyle \:v(0)\:=\:60\)

Substitute into [1]: \(\displaystyle \:v(0)\;=\;-32(0)\,+\,C_1\;=\;60\;\;\Rightarrow\;\;C_1\:=\:60\)

. . Hence, the velocity function is: \(\displaystyle \:\fbox{v(t)\;=\;-32t\,+\,60}\)


We want the height function, \(\displaystyle h(t)\)

Integrate: \(\displaystyle \:h(t) \;=\;\int v(t)\,dt \:=\:\int(-32t\,+\,60)\,dt \:=\:-16t^2\,+\,60t\,+\,C_2\;\) [2]

We are told that the ball starts on the ground: \(\displaystyle h(0)\,=\,0\)

Substitute into [2]: \(\displaystyle \:h(0) \;= \;-16(0^2)\,+\,60(0)\,+\,C_2\;=\;0\;\;\rightarrow\;\;C_2\,=\,0\)

. . Hence, the height function is: \(\displaystyle \:\fbox{h(t)\;=\;-16t^2\,+\,60t}\)


And now you can apply your techniques on these functions.

To find maximum height, differentiate the height function, equate to 0, and solve.
. . \(\displaystyle h'(t)\:=\:-32t\,+\,60\:=\:0\;\;\Rightarrow\;\;t \,=\,\frac{15}{8}\)

Then: \(\displaystyle \:h\left(\frac{15}{8}\right) \;=\;-16\left(\frac{15}{8}\right)^2\,+\,60\left(\frac{15}{8}\right) \;= \;\frac{225}{4}\;=\;\fbox{56.25\text{ ft}}\)