Anonymous

# Suppose a ball is thrown upward from a height of 5 ft with an initial velocity of 30 ft/sec.?

Suppose a ball is thrown upward from a height of 5 ft with an initial velocity of 30 ft/sec.?

I do I find the following?

a) Find the heigh of the ball after 1.5 seconds.

b) Find the max height reached by the ball.

c) Find the time it takes the ball to reach max height.

d) Find the time it takes the ball to hit the ground.

Relevance

The equation you would use is :

A= -16t^2+vt+h

With v equalling the initial velocity (30 ft/sec)

h equalling the initial height (5 feet)

t equalling time (for a:1.5 sec)

and A equalling the future height

So the equation would be A= -16t^2+30t+5

Then you should plug that into your graphing calculator and adjust your window as necessary. (Xmin:0, Xmax:3 Ymin:-10, Ymax:20 is a good window for this) I'm just going to assume you have a TI-83 or 84, because that is what have and it seems to be a pretty popular brand. If not, I'm sorry, but I'm sure your calculator can perform the same type of calculations for you.

For:

a) Find the y value when x=1.5 by using the Trace function (Press Trace then enter 1.5 and press the enter key).

b) Go to calc (2nd Trace) then go to maximum (#4) and choose your bounds around the peak of the parabola. The y-value it gives you is the max height and...

c) The x-value it gives you is the time it took to get to the max height.

d) Go to calc again and select zero (#2). You want to find the x-value where the parabola crosses the x-axis. (Your answer should be positive)

Good luck!

Source(s): Years of doing these types of problems...

hey Hardrockangel19!

a) d = do + vt + 0.5at^2, do=5', v=30ft/s, a=-g where g is the acceleration due to gravity (9.8m/s - convert to English units)

Therefore: d = do + vt - 0.5gt^2

b) set: do + vt - 0.5gt^2 = 0, and take the derivative and solve for t. therefore t=v/g. plug that into d = do + vt - 0.5gt^2 and solve for d

c) from b: t=v/g

d) using d = do + vt - 0.5gt^2, solve for t with d=0

hope this helps! email or IM me if it is not clear.

• heusel
Lv 4
4 years ago

Whenever you spot the phrase "highest", you must understand to discover the spinoff, and the significant aspects h'(t) = -32t + fifty two zero = -32t + fifty two 32t = fifty two t = fifty two/32 So there is a highest whilst t = fifty two/32 Plug that into the customary equation to discover the function at t = fifty two/32 h(fifty two/32) = -sixteen(fifty two/32)^two + fifty two(fifty two/32) = forty two.25 toes <----------------ANSWER