Homework help (Im stuck on this problem and have no clue on what to do)

Question

Dave throws a ball upward with an initial velocity of 32 ft/s. The ball initially leaves his hand 5 feet above the ground and eventually falls back to the ground. 

a. Using the model on the left ( h(t)=-16t^{2} +v{0}t+h{0} ), what would be the equation for the ball that Dave threw?

b. What is the maximum point reached by the ball? After how many seconds will it reach that height? 

c. How long will it take for the ball to land on the ground after being thrown? 

d. Label the following key features of the model and state what each tells you about the flight of the ball.

Vertex means-

X intercept means-

Y intercept means-

Answer 1

Question 1:

First, lets remember the standard form for a projectile, which you gave:

h(t) = -16t²+ v(0)*t + h(0) => We know that v(0) = 32 ft/s, and h(0) = 5ft

Our Equation becomes:

Answer to Question 1: h(t) = -16t² + 32t +5

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Question 2:

This equation is not easily factored, but we know that the vertex of the parabola will give us the best information. Since we know this is a projectile motion, we can assume an up-down facing parabola, where the vertex represents the maximum value of the height, at given time t.

Knowing that his is an up-down parabola, we know there is an axis of symmetry at t_v = - (b/2a) = - (32/[2*-16]) = 1. We can use time, t_v = 1 in our question to find that the height, h, at time t_v is: h(t_v) = -16(1)² + 32(1) +5 = 21 ft. This means that our parabola as a vertex at the point (1, 21).

Answer to Question 2:

Since this parabola faces down, we can safely assume that the maximum height for our ball is h_v = 21 ft, at time t_v =1 second.

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Question 3:

Here we know that h = 0 at the point where the ball hits the ground. Thus, we move into solving the regular ol' equation:

-16t² + 32t +5 = 0

Using the quadratic formula, you would come up with two answers:

t = 1 + Sqrt(21)/4.

Answer to question 3:

Since one of those values is negative (1 - sqrt(21)/4 = -0.14), you can assume that t = 1+ Sqrt(21)/4 = 2.14 seconds is the answer

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Question 4:

Here, you will simply need some graph paper, but you know the vertex is the maximum height, at time t = 1s (as described above). The X-Intercepts are the values of t we just calculated in Question 3. The left-most value really means nothing, physically. The right-most value is the time where the ball hits the ground after being thrown. You know that the Y-intercept is actually the height at time t = 0, or h(0), which was given to you as 5 ft.

Answer 2

height=Vo*t+a/2*t^2.

drop=(t^2+t)/2*32.2