Question 24750: 3) 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.
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? What is the maximum height of the ball? What time will the maximum height be attained?
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! a) s(t) = -16t^2 + 64t
b) s(1) = -16(1)^2 + 64(1) = 48 feet
c) s(t) = 0 at ground level, so
-16t^2 + 64 t= 0
Solve by factoring the common factor of -16t:
-16t ( t - 4) = 0
t= 0 or t= 4
Time to reach the ground is from t= 0 to t=4 which is 4 seconds.
Maximum height will be halfway between t=0 and t=4, which is t= 2 seconds to reach the maximum height. Maximum height will be
s(t) = -16t^2 + 64 t
s(2) = -16(2)^2 + 64(2)
s(2) = -64 + 128 = 64 feet
R^2 at SCC
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