# Cart project due 9/19 Wed. Check your rubric for expectations and grading procedure!! Contests fastest cart - 5 pts extra credit longest distance – 5 pts.

## Presentation on theme: "Cart project due 9/19 Wed. Check your rubric for expectations and grading procedure!! Contests fastest cart - 5 pts extra credit longest distance – 5 pts."— Presentation transcript:

Cart project due 9/19 Wed. Check your rubric for expectations and grading procedure!! Contests fastest cart - 5 pts extra credit longest distance – 5 pts extra credit prettiest – 5 stamps ugliest? - 2 stamps

Warm-up: verbal The formula I gave you for free fall equation #1a was a = vf –vi or a = ∆v t ∆t and vi = 0, what equation from your print out is the same as above vf = g∆t

Free Fall -- an Object Dropped Initial velocity is zero Final velocity  zero a g = -9.8 m/s 2. V f will be negative because it is moving in the negative direction. v i = 0 y x

Free Fall -- an Object Thrown Downward V i  zero because you apply a velocity to the object V f  zero. a g = -9.8 m/s 2

An object thrown upward Break up into two trajectories: Up and Down Up: – V i  zero because you apply a velocity to the object to make it travel up. – V f = zero v = 0 Down –V i = zero –V f  zero Remember that a g = -9.8 m/s 2

Equations for free fall, with constant acceleration of g when vi ≠ 0 when the vi = 0 vf = vi + g ∆t vf = g ∆t vf 2 = vi 2 + 2 g d vf 2 = 2 g d d = vi ∆t + 0.5 g ∆t 2 d = 0.5 g ∆t 2 d = ( vi + vf ) ∆t d = ( vf ) ∆t 2 2

1.A brick falls freely from a high scaffold at a construction site. a.What is its velocity after 4.0 seconds? Vi = 0, a = -9.8 t=4s vf = ? a= ∆v also vf = g ∆t (both will work!) ∆t

1b. How far does the brick fall during 4s? vi = 0, vf = -39.2 ms a = -9.8ms 2 t= 4s d = ? d = 0.5 g ∆t 2

2 a. A weather balloon is floating at a constant height above Earth when it releases a pack of instruments. If the pack hits the ground with a velocity of -73.5 m/s, how far does the pack fall? Vi =0, vf = -73.5 m/s a = -9.8 m/s 2 vf 2 = 2 g d we are using this equation because we do not have time!

2b. How long does the pack fall? vi = 0, vf = -73.5 m/s a = -9.8m/s 2 d = 276 m vf = g ∆t g = g

3a. A tennis ball is thrown straight up with an initial speed of 22.5m/s. It is caught at the same distance above the ground from which it was thrown. How high does the ball rise? vi = 22.5 m/s vf = 0 a = -9.8m/s 2 d=? Vf 2 = vi 2 + 2 g d remember a = g !!!

3b. How long does it take to reach that height? vi = 22.5 ms, vf = 0, a = -9.8m/s 2, d = 25.8m t=? Vf = vi + g ∆t

4a. A rock is thrown vertically upward at t = 0s It reaches a maximum height of 14m above the release point. What is the rock’s initial speed (velocity)? d = 14m a = -9.8 vf = 0 vi =? Vf 2 = vi 2 + 2 g d

4b. How long does it take to reach that height? d = 14 m, a = -9.8m/s 2, vf = 0, vi =17ms t = ? vf = vi + g ∆t

5a. A baseball is hit straight up in the air with an initial velocity of 38m/s. How long does it stay in the air? vi = 38m/s, a = -9.8m/s 2, vf = 0 ms, t =? Vf = vi + g ∆t 5b. What is the total distance the ball travels? d = vi ∆t + 0.5 g ∆t 2

a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved.

A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up.

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