Tutor profile: Sanjeem A.
A player kicks a soccer ball that travels 200 feet horizontally. The ball reaches an elevation of 60 feet. How high was the ball when it was 150 feet away from the student?
this is a parabola problem as the ball would travel in a parabolic path due to gravity. we know the equation of a parabola is f(x) = -a(x-h)^2 + k. we know that the highest elevation of the ball would be at the halfway point in the horizontal direction, so x and y would be 100 and 60 respectively. This point is also known as the vertex and the x and y points here are the h and k of the equation respectively. So plugging those 2 values into the equation, we have f(x) = -a(x-100)^2+ 60. We also know two more points, because the elevation at the beginning and end equals 0. so, the other two points are (0,0) and (160,0). We can use either points to solve for a. f(0) = 0 = -a(0-100)^2+60. -60 = -a(10000) a = 6/1000. so the equation is now f(x) = -6/1000(x-100)^2+60 plugging in x = 150, we get f(15) = -6/1000(150-100)^2+60 = (-6/1000)(50^2)+60 = -15+60 = 45 feet. Therefore, the ball will be 45 feet high when it is 150 feet away from the player.
A rectangle has a length that is 4 times it's width. If you halve the length and increase the width by 5 units, you still keep the same area. What is the area of the rectangle.
Area of a rectangle is length times width. if we consider the width of the original rectangle to be W, then the length is 4W. therefore the area of the original rectangle is (4W)(W) = 4W^2. The second rectangle has the Width 5 units longer, so the width in relation to the original rectangle is W+5. The length is half of the original rectangle, so it is 4W/2 = 2W. Therefore the area of the second rectangle is (W+5)(2W) = 2W^2+10W. We know that the area of the two rectangles is the same, so we can set the two equations to be equal to each other. 4W^2 = 2W^2 + 10W. 2W^2 - 10W = 0. W^2-5W = 0. W(W-5) = 0. W = 0 or 5. Obviously we cannot have W = 0 so the Width of the first rectangle is 5, and the length is 4x5 = 20. The area can be calculated by multiplying length x width = 20 x 5 = 100.
A train leaves a station at 100 miles per hour speed at 2 p.m. Another train leaves the same station on parallel tracks at 5:30 p.m. at 150 miles per hour. At what time and how far from the origin would the trains be at the same place?
We are solving for distance, which is a product of time and velocity. When the trains are at the same location, the distance traveled by both trains are the same. Assume time after 2 p.m. to be x hours. Therefore the distance traveled by the first train at any given time is 100(x) miles. Since the second train starts 2.5 hours later, the time traveled by the second train in relation to the first is x-2.5 hours. The distance traveled by the second train is therefore, 150(x-2.5) miles. Setting the two distances equal to each other, we get 100(x) = 150(x-2.5). Solving for x, we get, 100x = 150x-375. 50x = 375. Therefore x = 7.5. So the time they would be at the same place is 7.5 hours after 2 p.m., which is 9:30 p.m. The distance where they would meet can be solved by using x in either of the two equations, so, 100(7.5) = 750 miles from the origin.
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