Free Fall Examples

Example 2-14 Falling from a tower (v 0 = 0) Note! Take y as positive DOWNWARD! v = at y = (½)at 2 a = g = 9.8 m/s 2

Example 2-14 Falling from a tower (v 0 = 0) Note! Take y as positive DOWNWARD! v = at y = (½)at 2 a = g = 9.8 m/s 2

Example 2-14 Falling from a tower (v 0 = 0) v 1 = (9.8)(1) = 9.8 m/s v 2 = (9.8)(2) = 19.6 m/s v 3 = (9.8)(3) = 29.4 m/s Note! Take y as positive DOWNWARD! v = at y = (½)at 2 a = g = 9.8 m/s 2

Example 2-15: Thrown Down From a Tower A ball is thrown downward with initial velocity v 0 = 3.0 m/s, instead of being dropped. (a) Position after t = 1.0 s & 2.0 s? (b) Speed after t = 1.0 s & 2.0 s? Compare with speeds of a dropped ball. Photo of the leaning tower of Pisa

Example 2-15: Thrown Down From a Tower A ball is thrown downward with initial velocity v 0 = 3.0 m/s, instead of being dropped. (a) Position after t = 1.0 s & 2.0 s? y = v 0 t + (½)at 2 t = 1.0 s; y = (3)(1) + (½)(9.8)(1) 2 = 7.9 m t = 2.0s ; y = (3)(2) + (½)(9.8)(2) 2 = 25.6 m Photo of the leaning tower of Pisa

Example 2-15: Thrown Down From a Tower A ball is thrown downward with initial velocity v 0 = 3.0 m/s, instead of being dropped. (a) Position after t = 1.0 s & 2.0 s? y = v 0 t + (½)at 2 t = 1.0 s; y = (3)(1) + (½)(9.8)(1) 2 = 7.9 m t = 2.0s ; y = (3)(2) + (½)(9.8)(2) 2 = 25.6 m (b) Speed after t = 1.0 s & 2.0 s? v = v 0 + at t = 1.0 s; v = 3 + (9.8)(1) = 12.8m/s t = 2.0s ; v = 3 + (9.8)(2) = 22.6m/s Compare with speeds of a dropped ball. Photo of the leaning tower of Pisa

Examples 2-16, 2-18, 2-19 v A = v 0 = 15 m/s choose y as positive upward  a = -g = m/s 2 Time to the top = (½) round trip time! v = 0 here but still have a = -g v C = -v A (= -v 0 ) A person throws a ball up into the air with initial velocity v 0 = 15.0 m/s. Questions: a. Time to the top? b. Round trip time? c. Maximum height? d. Velocity when it comes back to the start? e. Times when the height y = 8.0 m?

Example 2-19: Ball thrown upward; the quadratic formula. For a ball thrown upward at an initial speed of v 0 = 15.0 m/s, calculate the times t the ball passes a point y = 8.0 m above the person’s hand.

Example 2-20: Ball thrown upward at the edge of a cliff. A ball is thrown up with initial velocity v 0 = 15.0 m/s, by a person standing on the edge of a cliff, so that it can fall to the base of the cliff 50.0 m below. Calculate: a. The time it takes the ball to reach the base of the cliff. b. The total distance traveled by the ball.

A stone is thrown at point (A) from the top of a building with initial velocity v 0 = 19.2 m/s straight up. The building is H = 49.8 m high, & the stone just misses the edge of the roof on its way down, as in the figure. Calculate: a) The time at which it reaches its maximum height. b) It’s maximum height above the rooftop. c) The time at which it returns to the thrower’s hand. d) It’s velocity when it returns to the thrower’s hand. e) It’s velocity & position at time t = 5 s. Example: Not a bad throw for a rookie!