(a) 1. Replace V0 = 56 ft/s and h = 40 ft in the equation h = -16t2 + V0t
40 = -16t2 + 56t
2. We have a quadratic equation that is solved as follows:
-16t2 + 56t -40 = 0
(-b ± √ (b ^ 2-4ac)) / 2aa = -16b = 56c = -40
t1 = 1.0 st2 = 2.5 s
Now we know that the ball initially reaches a height of 40 ft at 1 seconds and 2 seconds.
(b) 1. Replace V0 = 56 ft/s and h = 77 ft in the equation h = -16t2 + V0t
77 = -16t2 + 56t-16t2 + 56t -77 = 0
(-b ± √ (b ^ 2-4ac)) / 2a (quadratic equation)
When we replace the values in this equation, we obtain a negative root, which indicates that the ball never reaches a height of 77 ft.
(c) and (d):
When the ball is thrown straight upwards, there comes a point where it stops and starts to fall. That point where it stops is the highest point of its path(maximum height) and the speed is zero:
1. We use the formula of the final speed to find the time in which it reaches the maximum height:
V = at + V0 a = acceleration of gravity = 32 ft/s2
2. As V = 0, we have:0 = at + V0t = V0 / a
t = (56 ft/s) / 32 ft/s2
t = 1.75 s
3. We know that the ball reaches the maximum height at 1.75 seconds, so we substitute the value of "t" in the initial formula:
hmax = -16 (1.75) 2 + 56 (1.75)
hmax = 49 ft
(e) The time it takes to hit the ground will be when h = 0 ft, so we use it in the initial formula to find "t":
0=-16t2+56tt = 3.5 s
