Please help me complete my home assignment!
Answers
Answer:
(1) It should be possible to define the unit without ambiguity. (2) The unit should be reproducible. (3) The value of units should not change with space and time
h(t) = -16t2 + 108t + 28 (a) Maximum height occurs at the vertex of the height-vs.-time parabola, which is at t = -108/[2(-16)] sec = ? sec
Evaluate h(t) at this value of t to get h max.
. (b) Set h(t) = 0 and solve the quadratic equation for t. You will get a positive and a negative solution. Discard the negative solution, since time starts at t = 0.
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Total Answer!
As the given equation is equation of parabóla and the maximum height is at the vertex of the graph for the equation of parabola.
For finding the vertex of parabola is we need it's x and y coordinates.
x-coordinate of vertex is the time for maximum for maximum height while y-coordinate is the maximum height.
formula for x-coordin@te is -b/2a
comparing the given equation with the original equation of parabola we get the values of a, b and c as;
a=-16, b=108, and c=28
Here, x-coordinate is t=-b/2a
t=-108/(2*-16)
t=-108/(-32)
t=3.345 s
for y-coordinate (h), put the vàlue øf x-coordinate (i.e. value of t) in the given equation of parabola as;
h=-16(3.345)^2+108*3.345+28
h= -179.02+361.26+28
h=210.24 feet
Maximum height = h= 210.24 feet
Part.B How many seconds does it take uñtil the ball finally hits the grouñd?
If ball hits the ground it means y-component is zero.
putting y=O in the given equation we get the time that the ball will take to hit the ground
0=− 16 t^2 + 108 t + 28
using quadratic formul@, we get
pictures. the Continuation.
time cannot be negative so t=44 s.
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Done.
