Question

A stone is thrown upward from ground level. The initial speed is 48 feet per second. How high will it go?
Without using integrals lol.

Answer 1

s = ut + 0.5at^2
48(1) + 0.5(-10)(1)
48 + 5 = 43
assuming g = -10

Answer 2

why is g= -10?

Answer 3

@miteshchvm
need to use -32.2ft/s/s because units are in feet.

Answer 4

ya -32.2
you need to use g = -32.2 as it is going against gravity

Answer 5

Kinematics . . .
\[y_{final}=y_{initial}+v_{initial}(t)-16.1(t)^2\]

Answer 6

i cant tell which is which. :(

Answer 7

*from the problem

Answer 8

If you want some practice with your calculus, take that equation and differentiate once with respect to 't' to get the velocity, and differentiate again to get acceleration.

Answer 9

"A stone is thrown upward" = initial velocity is positive. ("The initial speed is 48 feet per second.")
"from ground level." = initial height = 0.
"How high will it go?" = solve for y_final.

Answer 10

how do i find t? or is it the same as speed..?

Answer 11

There are probably some other simplifying assumptions you will need.
One: neglecting air resistance, the parabolic trajectories of objects in free-fall are symmetric, which means they take the same amount of time to go up as they do to come down.

Answer 12

@cliffsedge
what is y final?
\[y_{final}=y_{initial}+v_{initial}(t)-16.1(t)^2 \]

Answer 13

You can also differentiate the position equation to get the velocity equation and realize that at the top of the flight, velocity =0.

Answer 14

@CliffSedge i guess you have written a wrong equation.

Answer 15

That is never the wrong equation. It is the fundamental kinematics equation, all others are derived from it.
Try this: take that equation, set it equal to zero for y_final and solve for t (use quadratic formula).

Answer 16

Granted, there are short-cut equations you can use for this sort of thing, but understanding where they come from will help out in the long run.

Answer 17

When you get more experience with this sort of thing, you'll see that if you differentiate the position equation:
\[\frac{d}{dx}(\Delta y = vt+0.5at^2) \rightarrow \Delta v = at\]
It becomes the velocity equation; if you solve it for t and substitute that in to the position equation, you'll get
\[\Delta (v^2)=2a \Delta y,\]
which lets you solve things without needing to know the time variable.