Question 77726: Please help
Suppose a baseball is shot up from the ground straight up with an initial velocity of 32 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 ½g, the gravitational pull due to gravity (measured in feet per second 2).
• 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.
a) What is the function that describes this problem?
b) The ball will be how high above the ground after 1 second?
c) How long will it take to hit the ground?
d) What is the maximum height of the ball?
Answer by renevencer22(21)   (Show Source):
You can put this solution on YOUR website!
a) this function describes displacement as, s = Vo*t - (1/2)*g*t^2, showing effect of gravity in motion of (falling and going-up ) body. It is in terms of initial velocity and gravity at any time.
see also: http://www.freewebs.com/renevencer11/-%20math/freefalling.jpg
note: copy, paste on your file & Ctrl+ click to view the link
b) after 1 sec s = Vo*t - (1/2)*g*t^2
Vo = 32 ft/sec s = 32*1 - (1/2)*32.2*1^2 = 15.90 ft
d) maximum height
V1^2 - Vo^2 = -2*g*h V1 = final velocity, equal to zero at maximum height
h = Vo^2/(2*g) = 15.90 ft
c) it becomes a free-falling body
s = Vo*t + (1/2)*g*t^2 V0 = 0
15.90 = (1/2)*32.2*t^2 =
t = sqrt(15.9 /(16.1))= 0.9938 sec
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