A football is thrown with an initial upward velocity component of 15.0 m/s and a horizontal velocity component of 18.0 m/s
The time required for the football to reach the highest point in its trajectory is 1.53s.
It gets 11.5m above the ground.
How much time after it is thrown does it take to return to its original height?
How far has the football traveled horizontally from its original position?
you are given time to get to top, it takes the same time to fall down.
how far=horizontal velocity*timeinair
1.53 seconds to get to highest point so 1.53x2 to reach the end of the porabola. 3.06 seconds
Jade you got to remember that it is accelerating downwards since gravity is -9.8 m/s. so it is not easy as that and I would think before you try and put your knowledge on the page.
Actually, Jacob, jade is 100% right as it is taught this way to save time but you can go the long way if you would like.
To find the time it takes for the football to return to its original height, we can use the fact that the total flight time is twice the time to reach the highest point.
Since the football takes 1.53 seconds to reach the highest point, the total flight time (time to reach highest point + time to fall back down) is 2 * 1.53 = 3.06 seconds. Therefore, after it is thrown, it takes 3.06 seconds for the football to return to its original height.
To find the horizontal distance traveled by the football, we can use the formula: distance = speed * time.
The horizontal velocity component of the football is 18.0 m/s. The total time of flight is 3.06 seconds.
Using the formula, distance = speed * time, we can calculate the horizontal distance traveled by the football: distance = 18.0 m/s * 3.06 s = 55.08 meters.
Therefore, the football has traveled 55.08 meters horizontally from its original position.