Suppose you throw a baseball straight up at a velocity of 64 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t2 + v0t + s0
•16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second2).
•v0 is the initial velocity (how hard do you throw the object, measured in feet per second).
•s0 is the initial distance above ground (in feet). If you are standing on the ground, then s0 = 0.
2006-06-17 03:24:57 · 4 answers · asked by jenniffer m 2 in Science & Mathematics ➔ Mathematics
Well, the initial velocity (v0) is 64 ft/s. Thus the equation becomes s = -16t^2 + 64 t + 0 . So, if you want the height after 1 s, just plug in 1 in the place of t. Thus height after 1s is -16 + 64 = 48 feet.
2006-06-17 03:35:23 · answer #1 · answered by organicchem 5 · 6⤊ 2⤋
s = -16t^2 + 64t
s = -16(1)^2 + 64(1)
s = -16 + 64
s = 48ft
2006-06-17 10:35:19 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
well, as you said the fuction is:
s=-16t^2+v0t+s0=-16t^2+64t.
plug 1 in for t.
2006-06-17 10:30:33 · answer #3 · answered by Eulercrosser 4 · 0⤊ 0⤋
that is not a real equation
2006-06-17 10:28:57 · answer #4 · answered by mab_209_kitty 1 · 0⤊ 0⤋