Parabolic Motion -04- Horizontal Distance In Parabolic Motion

Rebiaz Studio

A ball is fired horizontally with velocity of 50 m/s from the top of a hill 125 m high. Find the distance of the point where the particle hits the ground from foot of the hill.

Transcript

00:00Hi friends, there is always something new to learn and experience every Friday.

00:08Today we will learn the following question.

00:11A ball is kicked horizontally from the roof of a tall building.

00:15The ball will eventually touch the ground.

00:18Well, we are asked to calculate the horizontal distance of the ball from the ground floor

00:24of the building.

00:29Let's reconstruct this problem.

00:32A tall building is a building above ground level.

00:37As a reference, the ground surface is the x-axis.

00:41Thus the height value is always positive.

00:46The height of the building is 125 meters.

00:51The ball is kicked horizontally from point 0.

00:55This means that the initial velocity direction is in line with the x-axis.

01:01In other words, the initial elevation angle is 0 degrees.

01:08The ball will travel a parabolic path.

01:13Let's assume the ball reaches the ground at point 1.

01:19This is the distance we want to calculate.

01:24When solving the problem, we must analyze the quantities that have been stated on the

01:28problem sheet.

01:31The position vector at point 0, the vector S0, is 125 j-hat.

01:39The initial velocity vector is 50 i-hat.

01:43The position vector at point 1, the vector S, is Sx i-hat.

01:50Along the path, the ball will experience a downward vertical acceleration.

01:55The acceleration vector is minus 10 j-hat.

02:00Now write the kinematic equation for an object moving with constant acceleration.

02:06The vector S is equal to the vector S0 plus the vector V0t plus half the vector A, t-squared.

02:17Just plug in all the known values.

02:22The component vectors can only be added to values that have the same unit vector.

02:30Notice that there is no j-hat value on the left side.

02:33While on the right, there is a j-hat value.

02:38That is, 0 is equal to 125 5t-squared.

02:44From here, t-squared is equal to 25.

02:48Or t is equal to 5 seconds.

02:52We have found the value of t.

02:55Looking back at the previous equation, Sx is equal to 50t.

03:02Just substitute t equals 5.

03:05Sx is equal to 250 meters.

03:10It turns out that the ball landed at a point quite far from the ground floor of the building.

03:17Happy learning everyone!

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