Throwing A ball: A ball is thrown vertically from ground level with an initial speed of 108 ft/s.
A. When will the ball hit the ground?
B. When will the ball be 180 ft above the ground?
h=-16t^2VT+H
I mean
h=-16t^2+Vt+h
V = Vo + g*Tr = 0 at max h.
Tr = -Vo/g = -108/-32 = 3.375 s. = Rise
time.
Tf = Tr = 3.375 s. = Fall time.
A. T = Tr+Tf
B. h = Vo*t + 0.5g*t^2 = 180 Ft.
108t - 16t^2 - 180 = 0.
-16t^2 + 108t - 180 = 0.
Use Quad. Formula.
t = 3.0 s.
To answer these questions, we will use the equation h = -16t^2 + vt + h0, where h is the height, t is the time in seconds, v is the initial velocity, and h0 is the initial height from the ground.
A. When will the ball hit the ground?
We know that the initial height h0 is 0, and the initial velocity v is 108 ft/s. When the ball hits the ground, the height h will also be 0. Plugging in these values into the equation, we get:
0 = -16t^2 + 108t + 0.
To solve for t, we need to factor or use the quadratic formula. In this case, it is easier to factor out t:
t(108 - 16t) = 0.
Setting each term equal to 0, we get:
t = 0, or 108 - 16t = 0.
Solving the second equation:
108 - 16t = 0
16t = 108
t = 6.75 seconds.
So, the ball will hit the ground after approximately 6.75 seconds.
B. When will the ball be 180 ft above the ground?
Now we need to find the time when the height h is 180 ft. We can set up the equation as follows:
180 = -16t^2 + 108t + 0.
Again, we can simplify this equation:
-16t^2 + 108t - 180 = 0.
To solve for t, we can either factor or use the quadratic formula. Factoring the equation is a bit more complicated in this case, so we will use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a).
Plugging in the values of a, b, and c into the formula:
t = (-(108) ± √((108)^2 - 4(-16)(-180)))/(2(-16)).
Simplifying:
t = (-108 ± √(11664 + 11520))/(-32)
t = (-108 ± √(23184))/(-32)
t = (-108 ± 152)/(-32).
Now we have two possible solutions for t:
t1 = (-108 + 152)/(-32) = 4/8 = 0.5 seconds.
t2 = (-108 - 152)/(-32) = -260/-32 = 8.125 seconds.
Since time can only be positive, we discard the negative value of t. Therefore, the ball will be 180 ft above the ground after approximately 0.5 seconds.
Note: Keep in mind that these calculations are based on assuming no air resistance and no other external factors affecting the motion of the ball.
To solve this problem, we can use the equation you provided:
h = -16t^2 + Vt + H
Where:
h = height of the ball at a given time (in this case, it can be the ground or 180 ft)
t = time (in seconds)
V = initial velocity (in ft/s)
H = initial height (in ft)
Now let's solve for Part A: When will the ball hit the ground?
We know that the height h will be zero when it hits the ground, so we can set h = 0 in the equation:
0 = -16t^2 + 108t + H
Since the ball is thrown from ground level, H = 0. So the equation becomes:
0 = -16t^2 + 108t
To solve for t, we can factor out a common factor of t:
0 = t(-16t + 108)
Setting each factor equal to zero gives us two possibilities:
t = 0 (ignore this solution since time cannot be negative)
-16t + 108 = 0
Solving the second equation:
-16t = -108
Dividing both sides by -16:
t = 6.75 seconds
Therefore, the ball will hit the ground after 6.75 seconds.
Now let's move on to Part B: When will the ball be 180 ft above the ground?
We can set h = 180 in the equation:
180 = -16t^2 + 108t
Rearranging the equation to get it in standard quadratic form:
16t^2 - 108t + 180 = 0
To solve for t, we can either factor this quadratic equation or apply the quadratic formula. In this case, factoring might be the easier route:
16t^2 - 108t + 180 = 0
Dividing each term by 4 to simplify:
4t^2 - 27t + 45 = 0
Factoring the quadratic equation:
(4t - 9)(t - 5) = 0
Setting each factor equal to zero:
4t - 9 = 0
t - 5 = 0
Solving both equations:
4t = 9
t = 9/4 = 2.25 seconds
t = 5 seconds
Therefore, the ball will be 180 ft above the ground twice, at 2.25 seconds and 5 seconds after it is thrown.
I hope this explanation helps you to understand how to solve this problem. Let me know if you have any further questions!