a tennis ball is shot vertically upward from a launcher with an initial velocity of 64 ft per second. When will the ball return to the ground?

Question

anonymous · Answer

where you given an initial equation
h(t) = ??

anonymous · Answer

what you are looking for is h(0)

anonymous · Answer

ahhhmmm.. that was the given question but there are choices...

anonymous · Answer

a. 4 seconds
b. 4.8 sec
c. 48 sec
d. 48.5 sec

anonymous · Answer

using the information in the previous problem , how far off the ground will the ball be after 1 sec

anonymous · Answer

what is 64/16...do not have cal next to me

whpalmer4 · Answer

Come on, 64/16 is 4, you don't need a calculator for that!

anonymous · Answer

so a. 4 sec is the answer?

anonymous · Answer

No it's the wrong question...(sarcasm)

anonymous · Answer

but where did you get the 16???

anonymous · Answer

thanks whpalmer.....I think so keneisha...:)

basically the formula is 
H(t) = -16T^2 + 64T

solve for H(0)

anonymous · Answer

From Physics we know that the ball will reach a height of h feet after t seconds where h and t are related at the formula, based on the 64 given...

anonymous · Answer

t is 0 at launch
and 4 when it hits the ground...

anonymous · Answer

using the information in the previous problem , how far off the ground will the ball be after 1 sec

anonymous · Answer

use the same formula and t = 1

anonymous · Answer

a. 4.8 ft
b. 48  ft
c. 48.5 ft
d. 485 ft

whpalmer4 · Answer

$$h(t) = v_0t - \frac{1}{2}gt^2$$$$v_0 = 64 ft/s$$$$g = 32 ft/s^2$$
$v_0t$ gives you the upward travel, $-\frac{1}{2}gt^2$ gives you the downward component due to gravity.

So you set h(t) = 0 and solve for the two values of t.  The first value (t = 0) is the launch, the second is the landing.

whpalmer4 · Answer

attachment

anonymous · Answer

waht will be the answer in the second question?

whpalmer4 · Answer

Evaluate h(1) with the formula.

anonymous · Answer

i'm not really good in math...

whpalmer4 · Answer

$$h(1) = 64(1) - \frac{1}{2}(32)(1)^2 = $$

whpalmer4 · Answer

No time like the present to work on changing that!

anonymous · Answer

48 ft

anonymous · Answer

thanks a lot for the both of you
you two are really good in mathematics

whpalmer4 · Answer

There you go!  And if you look at my graph, you'll see that at the point over the 1 on the x-axis, it's just about to 50 on the y-axis.