Question

Use the formula h = −16t2 + v0t. A ball is thrown straight upward at an initial speed of v0 = 56 ft/s. (a) When does the ball initially reach a height of 40 ft? (b) When does it reach a height of 77 ft? (c) What is the greatest height reached by the ball? (d) When does the ball reach the highest point of its path? (e) When does the ball hit the ground?

Answer 1

A) The ball reaches 40ft when 40 = -16t^2 + v0t So 40 = - 16t^2 + 56t -16t^2 + 56t - 40 = 0 So the roots of the quadratic equation is x1 = 1 and x2 = 2.5 To find when the ball gets to that height we substitute. So (-16 * 1) + (56 * 1) + 40 = 0. Hence at t = x1; I. e at t= 1s

B) To find when it gets to 77ft we use the same approach. 77 = -16t^2 + 56t -16t^2 + 56t - 77 = 0 The root of the equation doesn't exist hence the ball doesn't get to that height.

C) To calculate the greatest height we need to calculate the time it reach that height first so it occured when derivative equals 0. - 16t^2 + 56t. So -32t + 56 = 0 t = -56/-32 = 1.75s Substitute into original equation we have H = (-16 *1.75^2 )+ (56 * 1.75) = -49 + 98 = 49ft

D) Already answered from C. Answer is in 1.75s E) The ball hits the ground when -16t^2 + 56t = 0 -16t^2 + 56t = 0 t( -16t + 56) = 0 t = 0 or t = 56/16 = 3.5s

Answer 2

(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