Projectile motion: Let's suppose you throw a ball straight up with an initial speed of 50 feet per second from a height of feet.
a) Find the equation that describes the motion as a function of time.
My answer was x = 0 and y=-16t^2+50t+6
b) How long is the ball in the air?
This is the part I need help on. Can I put the y equation into my graphing calculator to solve for t?
c) Determine when the ball is at maximum height. Find the maximum height.
when is y = 0 (hits ground)?
0 = 16 t^2 -50 t - 6 (I multiplied both sides by -1)
0 = 8 t^2 -25 t -3
solve quadratic for t
t = [ 25 +/- sqrt (625 + 96) ] / 32
t = [25 +/- 26.8514]/16
use the + root the - root is before you threw it
t = 3.24 seconds
max height when velocity = 0
V = Vo -32 t
32 t = 50
t = 1.56 seconds
height = 6 + 50 (1.56) - 16 (1.56^2)
Thank You!!
To find the time when the ball is in the air, you can set the y-coordinate equation equal to zero and solve for t.
The y-coordinate equation is: y = -16t^2 + 50t + h (where h is the initial height)
In this case, you mentioned that the ball was thrown from a height of feet. Let's assume the initial height is 6 feet. Hence, the equation becomes: y = -16t^2 + 50t + 6.
To solve for t, you can input this equation into a graphing calculator and find the points where the graph intersects the x-axis (since y = 0 at those points).
Alternatively, you can use the quadratic formula to solve for t. The quadratic formula states:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the equation is y = -16t^2 + 50t + 6, so a = -16, b = 50, and c = 6. Substituting these values into the quadratic formula will provide you with the two possible times when the ball is at y = 0 (or in the air).
c) To determine when the ball is at maximum height, you need to find the time at which the y-coordinate reaches its maximum value.
In projectile motion, the y-coordinate follows a parabolic path and reaches its maximum at the vertex of the parabola. The x-coordinate of the vertex can be found using the formula:
t = -b / (2a)
In this case, the equation is y = -16t^2 + 50t + 6, so a = -16 and b = 50. Substituting these values into the formula will give you the time at which the ball reaches its maximum height.
To find the maximum height, substitute the time obtained from the previous step into the y-coordinate equation. This will give you the y-coordinate of the maximum point.
Note: The values of a, b, and c in the equations depend on the specific initial conditions given in the problem. Make sure to adjust the equations accordingly based on the given initial height and the equation you derived.