2.27 Free Fall Problems
If the object is dropped from rest, the initial velocity ("vi") is zero. This further simplifies the equations to these:
g = vf /t d = 1/2 gt^2 vf^2 = 2gd
Practice #3: Mrs. Vail is fishing off a St. Johns River bridge. She hooks a big one, and in her excitement, her gum drops out of her mouth and falls to the water below in 2.15 seconds. Neglecting the ffects of air resistance, answer the questions below:
A) How far did the gum fall? t = 2.15 s g = -9.81 m/s^2 d = ? d = 1/2gt^2 = ½(-9.81 m/s^2)(2.15 s)2 = -22.67336 m d = -22.7 m The negative indicates that the gum fell 22.7 meters. B) What was the impact velocity of the gum? Again, we are looking for the final velocity. t = 2.15 s g = -9.8 m/s/s Vf = ? Vf = gt = (-9.81 m/s^2)(2.15 s) = -21.0915 m/s Vf = -21.1 m/s The negative indicates a downward direction; the impact of the velocity is 21.1 m/s in a downward direction.
Practice #1: Little baby bird is in a nest that is up in a big oak tree. Mama bird flies back to hungry little baby bird with a fat, juicy worm. In his haste, little baby bird drops the worm and sadly watches it fall to the ground below in 0.83 seconds.
A) How high is the nest? (Neglect the effects of air resistance.) t = 0.83 s g = -9.81 m/s^2 d = ? d = 1/2gt^2 = ½(-9.81 m/s^2)(0.83 s)2 = -3.379 m d = -3.38 m The negative denotes a downwards direction, meaning it falls down 3.38 m; therefore, the nest is 3.38 m high. B) How fast is the worm traveling when it hits the ground? (Neglect the effects of air resistance.) You are looking for the final velocity; the original or initial velocity is zero. t = 0.83 s g = - 9.81 m/s/s Vf = ? Vf = gt = (-9.81 m/s^2)(0.83 s) = -8.1423 m/s Vf = -8.1 m/s Notice the negative sign only tells you that the worm is moving downward. Vf =8.1 m/s
Practice #2: The Washington Monument in our nation's capital is 555 ft high.
A) If 1 m = 3.28 ft, what is the height of the Washington Monument in meters? This is a simple conversion. h = 555 ft(1 m/3.28 ft) = 169.207 m h = 169 m B) Neglecting the effects of air resistance, how long would it take a penny dropped from rest to fall from the top of the monument to the ground below? d = -169.207 m Notice that I am not using the 169 m which was put into significant figures to answer the last problem. g = -9.81 m/s^2 t = ? t = sqrt(2d/g) = sqrt{[(2)(-169.207 m)]/-9.81 m/s^2} = 5.8734 s Notice that the negative sign of the distance is divided out by the negative sign of the acceleration - this gives you a positive sign for the time - which is perfect. t = 5.87 s C) Neglecting the effects of air resistance, what would be the speed of the penny as it "hit" the ground? g = -9.81 m/s^2 t = 5.8734 s Vf = ? Vf = gt = (-9.81 m/s^2)(5.8734 s) = 57.618 m/s Vf = -57.6 m/s
Practice #4: Rosie looks out her second story window to watch her brother JJ playing with a tennis ball. JJ tosses the ball straight up and then catches it as it falls toward the ground. Rosie opens the window and yells, "JJ, throw the ball to me!" JJ gets right under her window and throws the ball straight up. Rosie reaches out and catches the ball as it reaches its highest point. If JJ releases the ball 1.4 m above the ground, and Rosie catches it 0.55 seconds later, how high above the ground was Rosie's hand as she caught the ball? Neglect the effects of air resistance.
In this problem, we will act as if the ball was dropped and not thrown upward. The time to go up and the time to drop that same distance is the same: t = 0.55 s g = -9.81 m/s^2 d = ? d = 1/2gt^2 = ½(-9.81 m/s^2)(0.55 s)^2 = -1.48376 m This means that Rosie is 1.48 m above JJ's hand; now you must add in the height JJ's hand is above the ground: total distance = 1.4 m + 1.48 m = 2.88 m
When you substitute the specific acceleration due to gravity (g), the equations are as follows:
g = (vf - vo)/t d = vot + 1/2 gt^2 vf^2 = vo^2 + 2gd
The following equations will be helpful:
a = (vf - vo)/t d = vot + 1/2 at^2 vf^2 = vo^2 + 2ad a = (vf - vo)/t d = vot + 1/2 at^2 vf2 = vo^2 + 2ad
