Question 51239: 3) 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 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?
Answer:
b) The ball will be how high above the ground after 1 second?
Answer:
Show work in this space.
c) How long will it take to hit the ground?
Answer:
Show work in this space.
d) What is the maximum height of the ball? What time will the maximum height be attained?
Answer:
Show work in this space.
Answer by THANApHD(104)   (Show Source):
You can put this solution on YOUR website! Do u know the function describe the distance with its acceleration,
D=ut+1/2a(t)^2
{ u- intitial velocity; a -acceleration;t-travelled time;D-distance between its starting point and destination}
so we will substitude Your data
D = 32t+(1/2*-32*(t^2)(-32/s^2 is the acceleration as the object travells againt the gravity)
= 32t+-16t^2
(here we must make some assumptions like , neglect the air friction,energy waste)
So if the ball is shot up above the height of "c" from sea level,
then the height of the ball after t second above sea level is E then
E=32t-16t^2+s
So after 1 s the height will be,
E=32-16+s
=16+s
If it stop its motion at this height the it would return towards the ground.
So there fore the equation
D=ut+1/2a(t)^2
16+s= 1/2*32*t^2(height is 16+s and gravity would be its acceleration cuz the object travells towards the gravity's direction, remembr the vectors)
t = [(16+s)/16]^(1/2)
= (16+s)^(1/2)/4 (unless we don't know thwe value of s, we can't evaluvate the t)
wat is acceleration, it is the incresement of velocity every one second.Strongly speaking, difference in velocity every second.
so there for acceleration of a equals to
a=v-u/t (where u-initial velocity,v-final velocity)
so in this case
-32=0-32/t (as the object goes against the gravity, gravity would be negative regarding its direction,and the object will go in its direction, unless its velocity reduces to 0.this is the state of motionless.where it changes it travelling direction to the gravity's direction.therefore the object would travel for 1 second,against the gravity,then it would stop it motion,and comeback to earth )
t=1second.
If t is 1 then substitude the t in the distance formula
E=ut+1/2a(t)^2+s
E=16+s
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