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)
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.
2 answers
Thank You!!
1 answer