An object is thrown horizontally from the top of a hill and hits the ground after 2 seconds. Find the height of the hill where the object was thrown. Suppose g = 10 m/s².
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00:00Hi friends, happiness is a hot bath on a Sunday afternoon.
00:07The following question is not hot enough.
00:10Just to practice the ability to answer physics questions.
00:15A ball is kicked horizontally from a fairly high hill.
00:20The ball reaches the ground two seconds later.
00:23Now, we are asked to estimate the height of the hill.
00:29As an initial illustration, here is a hill.
00:35A ball is on top of the hill.
00:39The ball will move through a parabolic path.
00:45To analyze this process, place the system in two-dimensional Cartesian coordinates.
00:53The starting point of the ball is point 0, while the point of the ball on the ground is point 1.
01:02The initial height of the ball is S0y. This is the value we want to calculate.
01:09Note for a moment, the direction of the initial velocity is parallel to the x-axis.
01:15This means that the initial elevation angle is 0 degrees.
01:19Now write the kinematic equation for an object moving with constant acceleration.
01:24Vector S1 is equal to vector S0 plus vector V0t plus vector At squared.
01:32This is a vector equation. The vector has a vertical component and a horizontal component.
01:40The height of the ball is in the vertical component,
01:43so we will use the components in the vertical direction.
01:46At point 1, the height of the ball is 0.
01:52The ball has no initial velocity component in the vertical direction.
01:58The acceleration due to gravity is toward the negative y-axis.
02:04The time it takes for the ball to reach point 1 is 2 seconds.
02:08I think this is an easy calculation. S0y is equal to 20 meters.
02:14This is the estimate of the height of the hill. This is easy, isn't it? Happy learning everyone!