A baseball is thrown upward from ground level with an initial velocity of 48 feet per second, and its height h in feet is given by , where t is the time in seconds.
At what time does the ball reach 10 feet? Round to the nearest tenth.
Step by step please
h(t) = 48t - 16t^2
so, plug in 10 for h and solve for t:
10 = 48t - 16t^2
8t^2 - 24t + 5 = 0
t = (6±√26)/4
or
t=0.225s (on the way up)
t=2.774s (on the way back down)
To find the time when the ball reaches a height of 10 feet, we need to solve the equation for h = 10. The equation is h = -16t^2 + 48t, where h is the height of the baseball in feet and t is the time in seconds.
1. Substitute h = 10 into the equation:
10 = -16t^2 + 48t
2. Rearrange the equation:
16t^2 - 48t + 10 = 0
3. To solve this quadratic equation, we can either use factoring or the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For the given equation, a = 16, b = -48, and c = 10.
t = (-(-48) ± √((-48)^2 - 4(16)(10))) / (2(16))
= (48 ± √(2304 - 640)) / 32
= (48 ± √1664) / 32
4. Calculate the square root of 1664:
√1664 ≈ 40.796
5. Substitute the square root into the equation:
t = (48 ± 40.796) / 32
6. Find both solutions:
t1 = (48 + 40.796) / 32 ≈ 2.402
t2 = (48 - 40.796) / 32 ≈ 0.184
7. Round both solutions to the nearest tenth:
The ball reaches a height of 10 feet at approximately t = 2.4 seconds and t = 0.2 seconds (rounded to the nearest tenth).
Therefore, the ball reaches a height of 10 feet at approximately 2.4 seconds (rounded to the nearest tenth).
To find the time at which the ball reaches a height of 10 feet, we need to set up the equation and solve for t.
Given:
Initial velocity (u) = 48 feet per second
Height (h) = 10 feet
Time (t) in seconds
The equation that relates the height of the ball to time is given by:
h = ut - (1/2)gt^2
Where g is the acceleration due to gravity, which is approximately 32 feet per second squared.
Substituting the given values into the equation:
10 = (48)t - (1/2)(32)t^2
Simplifying this equation:
10 = 48t - 16t^2
Rearranging the equation to bring it to standard quadratic equation form:
16t^2 - 48t + 10 = 0
Now we can solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 16, b = -48, and c = 10.
Substituting the values into the formula:
t = (-(-48) ± √((-48)^2 - 4(16)(10))) / (2(16))
t = (48 ± √(2304 - 640)) / 32
t = (48 ± √1664) / 32
t = (48 ± 40.8) / 32
Now we have two possible solutions for t:
t1 = (48 + 40.8) / 32 = 88.8 / 32 = 2.775
t2 = (48 - 40.8) / 32 = 7.2 / 32 = 0.225
We discard the negative value, so the ball reaches a height of 10 feet approximately 2.8 seconds after being thrown.
Therefore, the ball reaches a height of 10 feet at approximately 2.8 seconds.