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Mechanics 5 Applying the SUVAT equations to solve problems in one and in two dimensions IFP 6th November 2015.

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Presentation on theme: "Mechanics 5 Applying the SUVAT equations to solve problems in one and in two dimensions IFP 6th November 2015."— Presentation transcript:

1 Mechanics 5 Applying the SUVAT equations to solve problems in one and in two dimensions
IFP 6th November 2015

2 Defining a projectile Examples?
Ball, stone, Anything thrown A projectile is a body which is acted upon only by gravity (normally we just forget about air resistance )

3 Analysing Projectile Motion
Back to resolving vectors! Horizontally - force acting? Vertically –force acting? So we just apply SUVAT separately for X and for Y.

4 Is horizontal motion really independent of Vertical?
Train and tennis ball Theory quick intro Walter Lewin Theory For home study if you don’t get it YES! So we can use SUVAT in the Horizontal direction and in the Vertical direction independently

5 Apply suvat to x (horizontal and y (vertical) separately
sx is horizontal displacement ux is initial horizontal velocity vx is final horizontal velocity u = v horizontally ax is acceleration in horizontal plane = 0 tx is time sy is vertical displacement uy is initial vertical velocity vy is final vertical velocity ay is acceleration in vertical plane = -9.8 ms-2 ty is time ty = tx

6 Apply suvat to x (horizontal and y (vertical) separately
sx is horizontal displacement ux is initial horizontal velocity vx is final horizontal velocity ax is acceleration in horizontal plane tx is time sy is vertical displacement uy is initial vertical velocity vy is final vertical velocity ay is acceleration in vertical plane ty is time

7 Setting out problems – what we know
X y sx is horizontal displacement ux is initial horizontal velocity vx is final horizontal velocity u = v horizontally ax is acceleration in horizontal plane = 0 tx is time sy is vertical displacement uy is initial vertical velocity vy is final vertical velocity ay is acceleration in vertical plane = -9.8 ms-2 ty is time ty = tx

8 Projectile question -2Dimensional SUVAT question
A tennis ball is struck such that it leaves the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance 19.6 m from the racket. What is the height at the tennis ball when it leaves the racket?

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10 Practice with projectile problems
(if you get these don’t bother with the next few slides really)

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12 Further Example 1 A ball thrown vertically upward is caught by the thrower after 2.0 s. Find the initial speed of the ball the maximum height the ball reaches

13 One way to do this…

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15 Example 2 (non zero angle)
A projectile is fired from ground level at an angle θ upward from the horizontal and a speed of u. How long was the projectile in flight? What was its range R? What was the maximum height H?

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