A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is
s (t) = 112 + 96t - 16t 2
112 + 96t - 16t^2= s (t)
112=112=96t-16t^2
96t-16t^2=0
t(96-16t) =0
96=16t
t=6
Replace s(t) with 112 ft in a bid to solve for (t) which is time
Minus the 112 from both ends of the equation thus we have the t using zero-product principle
Then we will factor out t
Time will be zero at the zero seconds when the ball when the ball in its original state which is 112
Delegate your assignment to our experts and they will do the rest.
Divide both ends using 16
Thus at the sixth second the ball will have returned to the original height which is 112ft
t=0.5 s(t)=112-96t-16t^2
112+96(0.5)-16(0.5)^2= s (t)
112+48-4= s (t)
156= s (t)
we will also replace here t with the o.5 ft in a bid to resolve for distance s(t)
96*0.5 also -16 with (0.5)^2
Then we will add and also minus the remaining numbers
s(t) = 156 seconds this is at 5 seconds while the height of the ball is 156ft
1 s (t)=112+96t-16t^2= s (t)
112+96(1)-16(1)^2= s (t)
112+96-16= s (t)
192= s (t)
We replace t with 1 ft in a bid to resolve for distance (s(t)
96*1 then -16*(1)^2
We add and minus the remaining numbers
s(t) = 192 at the second 1 the height of the ball will be 192 ft
2 s (t)=112+96t-16t^2
112+96(2)-16(2)^2= s (t)
112+192-64= s (t)
192= s (t)
We replace t again with 2 in a bid to resolve for distance t
96*2 then -16*(2)^2
We add and minus the remaining numbers
s(t)=240 that means that at the second second the ball’s height will be 240ft
and 6) s(t)=100 s(t) = 112 + 96t – 16t^2
100 = 112 + 96t – 16t^2
-12-96t + 16t^2 = 0
4t^2 – 24t – 12 = 0
t = -(-24) plus or negative square root of (-24)^2 – 4.4(-3)/2.4
t = 24 plus or negative square root of (-24)^2 – 4.4(-3)/2.4
t = 24 plus or negative square root of 576 + 48/8
t = 24 plus or negative 4 square root of 39/8
t = 24 plus 4 square root of 39/8
t = 24 negative 4 square root of 39/8
t1 = 6 negative square root of 39/2
t1 = 6 plus square root of 39/2
t1 = -0.1225(to four decimals)
t2 = 6.1225
and 8). s(t)=200 s(t) = 112 + 96t – 16t^2
200 = 112 + 96t – 16t^2
200 = 112 + 96t – 16t^2 = 0
88 – 96t + 16t^2 = 0
t = -(-12) plus or negative square root of (-12)^2 – 4.2(-11)/2.2
t = 12 plus or negative square root of (-12)^2 – 4.2(-11)/2.2
t = 12 plus or negative square root of 144 – 88/8
t = 12 plus or negative 2 square root of 14/4
t 1= 12 negative 2 square root of 14/4
t 2= 12 plus 2 square root of 14/4
t1 = 1.1292
t2 = 4.8708
For steps 5 and 6, s(t) is replaced with 100 and expression equated to zero. The equation is then shifted to the left side so that positive can be changed to negative and vice versa. The like terms are then subtracted and reordering done, then simplification follows. Factoring out the like terms and reducing by the fraction of 4 is done to get to the answer.
For steps 7 and 8, (t) for time is replaced with 200, making the expression to be equal to zero where the equation is shifted to the left so that negative can be changed to positive and vice versa. The like terms are then subtracted and reordering is done to help simplify with 8 as the common factor. Subtracting and simplifying the numbers is done then reducing by a fraction of 2 to get the solution.
Complete the table and discuss the interpretation of each point.
| t | s(t) | Interpretation |
|---|---|---|
| 0 | 112. 0 |
Zero seconds which is represented by (t) represents the original height of the ball which is 112.0 feet Therefore, s(t) is represented by (0, 112.0) |
| 0.5 | 156.0 | 0.50 seconds (t) represents 156.0 feet as height of the ball, (0.5.0,156.0) |
| 1 | 192.0 | When (t) is one second, the ball reached 192.0 feet making (1, 192.0) |
| 2 | 240.0 | Two seconds represents the height of 240.0 feet which the ball reached |
| -0.1225 | 100 |
When the ball is at 100 feet, there are two y coordinates thus representing a Parabola graph. At 100 feet, the ball will have taken negative time which is -0.1225 seconds. It is an indication that the ball did not get thrown but it fell down at twelve feet. |
| 6.1225 | 100 |
The ball falling at 100 feet represents a Parabola graph due to the two varying y coordinates. The ball is thrown up and it takes 6.1225 seconds to get to the 100 feet. |
| 1.1292 | 200 | The height of the ball after it has been thrown up and descended is 200 feet. The time taken for the ball to go up and get down is 1.1292 seconds. |
| 4.8708 | 200 | The 200 feet measure is the height of throwing the ball up and descending down to the ground. The time taken for the 200 feet to be covered is 4.8708 seconds. |
Answer these questions.
After how many seconds does the ball strike the ground?
Seven seconds is the time the ball takes to get to the ground
112.0 + 97(7) – 16 (7) ^2 = s(t)
112.0 + 672 – 16.49 = s(t)
112.0 + 672 – 784 = s(t)
0= s(t)
After how many seconds will the ball pass the top of the building on its way down?
The ball takes six seconds to get to the top of the building as shown in the working in question one.
How long will it take the ball to reach the maximum height?
The maximum height is attained after three seconds.
112.0 + 96 (3) – 16(3) ^2= s(t)
112.0 + 288.0 – 16.90 = s(t)
112.0 + 288.0 – 144.0 = s(t)
256.0 feet= s(t)
What is the maximum height?
256.0 feet as shown in calculations in question three
There are is no way to get to the solution.
