An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is 50 ft per sec. The height s of the ball in feet is given by the equation s=-2.7t^2+50t+6.5, where t is the number of seconds after the ball was thrown. Complete parts a and b.
A.) After how many seconds is the ball 10ft above the moon's surface?
B.) how many seconds will it take for the ball to hit the moon's surface?
A.) To find out after how many seconds the ball is 10 feet above the moon's surface, we need to set the height equation equal to 10 and solve for t.
s = -2.7t^2 + 50t + 6.5
10 = -2.7t^2 + 50t + 6.5
We can rearrange the equation to form a quadratic equation:
-2.7t^2 + 50t + 6.5 - 10 = 0
-2.7t^2 + 50t - 3.5 = 0
Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -2.7, b = 50, and c = -3.5
t = (-50 ± √(50^2 - 4(-2.7)(-3.5))) / (2 * -2.7)
Solving this equation will give us two values of t. The positive value will represent the time when the ball is 10 feet above the moon's surface.
B.) To find out how many seconds it will take for the ball to hit the moon's surface, we need to set the height equation equal to 0 and solve for t.
s = -2.7t^2 + 50t + 6.5
0 = -2.7t^2 + 50t + 6.5
Again, we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -2.7, b = 50, and c = 6.5
t = (-50 ± √(50^2 - 4(-2.7)(6.5))) / (2 * -2.7)
Solving this equation will give us two values of t. The positive value will represent the time when the ball hits the moon's surface.
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To find the answers to parts A and B, we need to calculate the time (t) at which the height of the ball (s) is equal to the given values. Let's solve these step by step:
A.) After how many seconds is the ball 10 ft above the moon's surface?
We need to find the value of t when s = 10 ft, so we can substitute s = 10 into the equation:
10 = -2.7t^2 + 50t + 6.5
Now, we need to solve this quadratic equation for t. To solve it, we can rearrange it as follows:
2.7t^2 - 50t + 6.5 - 10 = 0
2.7t^2 - 50t - 3.5 = 0
Next, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Where a = 2.7, b = -50, and c = -3.5. Plugging in these values, we get:
t = (-(-50) ± √((-50)^2 - 4(2.7)(-3.5))) / (2(2.7))
t = (50 ± √(2500 + 37.8)) / 5.4
t = (50 ± √(2537.8)) / 5.4
Calculating the square root of 2537.8, we get:
t ≈ (50 ± 50.377) / 5.4
Now, we can calculate both possible values of t:
1. t ≈ (50 + 50.377) / 5.4
t ≈ 100.377 / 5.4
t ≈ 18.617 seconds
2. t ≈ (50 - 50.377) / 5.4
t ≈ -0.377 / 5.4
t ≈ -0.07 seconds
Since time cannot be negative in this context, the ball is approximately 10 ft above the moon's surface after 18.617 seconds.
B.) How many seconds will it take for the ball to hit the moon's surface?
To find the time it takes for the ball to hit the moon's surface, we need to find the value of t when s = 0 ft. We can substitute s = 0 into the equation:
0 = -2.7t^2 + 50t + 6.5
Similar to part A, let's solve this quadratic equation for t:
2.7t^2 - 50t - 3.5 = 0
Using the quadratic formula again, we can find the value of t:
t = (-b ± √(b^2 - 4ac)) / 2a
Where a = 2.7, b = -50, and c = -3.5. Plugging in these values:
t = (-(-50) ± √((-50)^2 - 4(2.7)(-3.5))) / (2(2.7))
t = (50 ± √(2500 + 37.8)) / 5.4
t = (50 ± √(2537.8)) / 5.4
Calculating the square root of 2537.8:
t ≈ (50 ± 50.377) / 5.4
Now, we can calculate both possible values of t:
1. t ≈ (50 + 50.377) / 5.4
t ≈ 100.377 / 5.4
t ≈ 18.617 seconds
2. t ≈ (50 - 50.377) / 5.4
t ≈ -0.377 / 5.4
t ≈ -0.07 seconds
Similar to part A, since time cannot be negative in this context, we can conclude that the ball hits the moon's surface after approximately 18.617 seconds.
A.) To find the number of seconds after which the ball is 10 ft above the moon's surface, we can substitute the given value of s into the equation and solve for t.
Given:
s = 10 ft
s = -2.7t^2 + 50t + 6.5
Substituting s = 10 in the equation:
10 = -2.7t^2 + 50t + 6.5
Rearranging the equation:
-2.7t^2 + 50t + 6.5 - 10 = 0
-2.7t^2 + 50t - 3.5 = 0
Now, we can solve this quadratic equation for t, either by factoring or by using the quadratic formula. Let's use the quadratic formula here:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values:
t = (-(50) ± √((50)^2 - 4(-2.7)(-3.5))) / (2(-2.7))
Simplifying:
t = (-50 ± √(2500 - 37.8)) / (-5.4)
t = (-50 ± √(2462.2)) / (-5.4)
Solving for t using a calculator or other methods:
t ≈ 2.8 seconds (rounded to one decimal place)
Therefore, after approximately 2.8 seconds, the ball will be 10 ft above the moon's surface.
B.) To find the number of seconds it takes for the ball to hit the moon's surface, we need to find the value of t when s (height) is 0.
Given:
s = -2.7t^2 + 50t + 6.5
Setting s = 0:
0 = -2.7t^2 + 50t + 6.5
We can solve this quadratic equation in a similar manner as in part A. Using the quadratic formula, we get:
t = (-(50) ± √((50)^2 - 4(-2.7)(6.5))) / (2(-2.7))
Simplifying:
t = (-50 ± √(2500 - 70.2)) / (-5.4)
t = (-50 ± √(2429.8)) / (-5.4)
Solving for t:
t ≈ -0.07 seconds (rounded to two decimal places) or t ≈ 18.89 seconds (rounded to two decimal places)
Since time cannot be negative in this context, we discard the negative solution. Therefore, it will take approximately 18.89 seconds (rounded to two decimal places) for the ball to hit the moon's surface.
a)
10 = -2.7t^2 + 50t + 6.5
2.7t^2 - 50t + 3.5 = 0
solve for t , using the quadratic formula, remember you have 2
positive answers,
both are valid, one is on the way up, the other on the way down
b) solve
-2.7t^2 + 50t + 6.5 = 0 , again using the quadratic formula