A ball is launched from a height of 6 feet at a speed of 24 feet/second. If the ball follows projectile motion, how long will it take for the ball to touch the ground?
(Hint: For projectile motion, height (h) = vt – 0.5gt^2 + initial launch height, and acceleration due to gravity (g) = 32 feet/second^2.)

Question

A ball is launched from a height of 6 feet at a speed of 24 feet/second. If the ball follows projectile motion, how long will it take for the ball to touch the ground?
(Hint: For projectile motion, height (h) = vt – 0.5gt^2 + initial launch height, and acceleration due to gravity (g) = 32 feet/second^2.)

anonymous · Answer

the choices are-
3.125 minutes
4.625 minutes 
3.125 seconds
1.718 seconds

jhannybean · Answer

Do you know the basic equation of this function?

anonymous · Answer

no, I wasn't given one.

jhannybean · Answer

Alright, projectile motion follows the formula $$\large y= \frac12 at^2+ v_{0}t+y_{0}$$where $y$ = height, $t$ = time, $a$ = acceleration due to gravity, $v_{0}$ = initial velocity ,$y_{0}$ = initial height

jhannybean · Answer

Just plug in all your variables and solve :)

anonymous · Answer

okay thank you, i just have one more question, what would be my initial height and initial velocity?

bahrom7893 · Answer

$$V_i=24\frac{ft}{sec}$$

jhannybean · Answer

JUST KIDDING. i didnt know it stated that in the problem.

bahrom7893 · Answer

No, you don't know any math and should stop helping on OS, since it only confuses the asker!

jhannybean · Answer

:(

bahrom7893 · Answer

:'(

anonymous · Answer

soo..... what?

jhannybean · Answer

BUT there WILL be a point, the climax of the ball's height that it WILL be 0, thats right after its launched and just before it falls.

bahrom7893 · Answer

not necessarily right after it launches ;)

anonymous · Answer

anyone want to let me know how to solve this problem?

bahrom7893 · Answer

But yes, there will, which can be shown here:
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