Question

# Equations of motion are used to describe the velocity (speed and direction) as well as distance...

Equations of motion are used to describe the velocity (speed and direction) as well as distance travelled by projectiles when they are launched or thrown. The following equation is used to model this model this movement: ℎ = −16?2 + ?0? + ℎ0

A ball is launched upward at 48 feet/second from a platform that is 100 feet high. Find the maximum height the ball reaches and how long it will take to get there. a) Substitute the appropriate values to create an equation of motion for this problem, using the general equation for motion provided in the “Introduction section” of this document. b) Find the vertex of the parabola that the ball will trace. What does this point represent in the path of motion? c) Find the x and y intercept of this parabola of motion (the equation you found above). Round your answers to the nearest hundredth value. d) Provide the domain and range of this equation. What do these values represent in our model? e) Create a graph that accurately represents this motion f) Find the height that the ball has reached at the following time values: i) ?? = 0.7 ?????????????? ii) ?? = 1.0 ?????????????? iii) ?? = 1.2 ?????????????? g) Find the height of the ball using the graph and your equation of motion when ?? = 1.0 ??????????????. Are these values exactly the same or are they different?

Explain why there may be a discrepancy in the two values. h) Do the values that you calculated seem reasonable for a ball free falling through the air? Explain using general logic and expectations.

given, h = -16t^2 + vo*t + ho

given, ball is launched upward,voy = 48 ft/s
initial platform height yo = 100 ft

a. max height be H
at max height vy = 0
now
vy = voy - gt
and
H = yo + voy*t - 0.5gt^2

so, vy = 0 ,
t = voy/g
H = yo + voy^2/2g = 100 + 48^2/2*32 = 136 ft

b. y = 100 + 48t - 16t^2
and x = 0
vertex of parabola will be (0,136 ft)
this represent the maximum height

c. y = yo + voy*t - 0.5gt^2 = 100 + 48t - 16t^2
vertex is (0,136)
x itercept 0
y intercept = 100 ft

d. domain is t > 0
range is 0 <= y <= 136 ft

e. the graph with respect to time is as under

f. h(0.7) = 100 + 48*0.7 - 16*0.7*0.7 = 125.76 ft
h(1) = 100 + 48*1 - 16*1*1 = 132 ft
h(1.2) = 100 + 48*1.2 - 16*1.2*1.2 = 134.56ft

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