ball is thrown up at 88 ft/s from a height of 25 feet. Path of ball is
y = -16t^2 + 88t +25. y is the height of ball above the ground t seconds after being thrown. when does it reach maxiumum height? (I got that at 2.75 s)
At what time does the ball hit the ground? (I set equation to =0, but I got 5.2 s and answer is 5.77 s)
Vf = Vo(t) - 32(t)
Vf = 88 - 32t or t = 2.75sec.
Height reached from 25 ft. height
h = 88(t) - 32(t^2)/2
h = 88(2.75) - 16(2.75)^2 = 121 ft.
It takes the same 2.75 sec. to return to the original launch height of 25 feet and an increment more to fall the 25 feet further to the ground.
Therefore,
h = Vo(t) + 32(t^2)/2
(121 + 25) = 0(t) + 16(t^2)
t - sqrt(146/16) = 3.02
The time from launch to impact is therefore (2.75 + 3.02) = 5.77 sec.
To find the time at which the ball reaches its maximum height, we need to find the vertex of the parabolic function y = -16t^2 + 88t + 25. The vertex of a parabola is given by the formula t = -b / (2a), where a, b, and c are the coefficients of the quadratic equation.
For our equation, a = -16 and b = 88. Plugging these values into the formula, we have:
t = -88 / (2 * -16) = -88 / -32 = 2.75 s
So, you are correct that the ball reaches its maximum height after 2.75 seconds.
To find the time at which the ball hits the ground, we need to set the equation to 0 and solve for t. Therefore, we have:
-16t^2 + 88t + 25 = 0
We can either solve this equation by factoring, completing the square, or using the quadratic formula. Since factoring may be difficult in this case, let's use the quadratic formula. The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
Applying this formula to our equation, we have:
t = (-88 ± √(88^2 - 4(-16)(25))) / (2(-16))
t = (-88 ± √(7744 + 1600)) / (-32)
t = (-88 ± √(9344)) / (-32)
t = (-88 ± 96.6) / (-32)
Using the positive square root since time cannot be negative in this context, we have:
t = (-88 + 96.6) / (-32)
t = 8.6 / -32
t ≈ -0.27
Since time cannot be negative, we disregard this negative solution.
Therefore, the ball hits the ground after approximately 0.27 seconds, or about 5.77 seconds, as you mentioned.
To find the maximum height and the time at which the ball reaches it, we need to determine the vertex of the parabolic equation y = -16t^2 + 88t + 25. The vertex represents the highest point of the parabolic curve, which in this case is the maximum height of the ball.
The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula:
t = -b / (2a)
In our case, the equation is y = -16t^2 + 88t + 25. The coefficient of t^2 is -16, and the coefficient of t is 88. Plugging these values into the formula, we get:
t = -88 / (2 * -16)
t = -88 / -32
t = 2.75 seconds
So, the ball reaches its maximum height at 2.75 seconds.
Now, let's find the time at which the ball hits the ground. To do that, we solve the equation y = -16t^2 + 88t + 25 for t when y = 0 since the ball hits the ground when its height is zero.
0 = -16t^2 + 88t + 25
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -16, b = 88, and c = 25. Plugging these values into the quadratic formula, we get:
t = (-88 ± √(88^2 - 4 * -16 * 25)) / (2 * -16)
Calculating further, we have:
t = (-88 ± √(7744 + 1600)) / -32
t = (-88 ± √(9344)) / -32
Now, we can simplify and calculate the two possible solutions:
t = (-88 + √9344) / -32 ≈ 5.77 seconds
t = (-88 - √9344) / -32 ≈ -0.18 seconds
Since time cannot be negative in this context, we discard the negative solution. Therefore, the ball hits the ground at approximately 5.77 seconds.
So, the time it takes for the ball to reach its maximum height is 2.75 seconds, and the time it takes for the ball to hit the ground is approximately 5.77 seconds.