Question

A ball is thrown into the air with an upward velocity of 48 ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+48t+6. How long does it take the ball to reach is maximum height? What is the balls maximum height? Round to the nearest 100th, if necessary.


A.) 1.5 s; 54 ft
B.) 1.5 s; 42 ft
C.) 3 s; 6 ft
D.) 1.5 s; 114 ft

Answer 1

Well, if you've done come calculus. we can take the derivative.

this function tells us the height as a function of time.

to find the maximum height we can take the derivative and set it equal to zero.

\[\frac{ dh }{ dt }(-16t^{2}+48t+6) = 0 \]

\[-32t+48 = 0 \]

\[t = \frac{ -48 }{ -32 } = 1.5 \]

you can plug this into the equation above

\[f(1.5) = -16(1.5)^{2}+48(1.5)+6 \]

the other way would be to consider that at the maximum height the velocity is equal to zero.

so let's start with the second method.

here's your physics equation for that. Ignore the math portion

look familiar?

\[y = x_{0}t + \frac{ 1 }{ 2 }at^{2}+v_{0}t \]

we know that the change in position over time = velocity

\[V(t) = \frac{ dx }{ dt }( x_{0}+2at+V_{0})\]

let's assume x_{0} = 0

that's how we get our equation for velocity

\[V(t) = V_{0}+at \]

V0 our initial velocity = 48ft/s
so we know at the highest point Vf = 0

\[0 = V_{0}-at \]

\[t = \frac{ V_{0} }{ a } = \frac{ 48 (ft/s) }{ 32.4 (ft/s^{2}) } = 1.5s \]

so, there you have two methods both that give the same time.

with this you should be able to get the rest of it.

Answer 2

Thank you :)

Answer 3

anytime but you see @Anna8122 whenever you throw something say into the air, the force of gravity is slowing the object until it gets to the top.