Question #3, p

Question #3, p. 35

On Mars, the acceleration due to gravity is -3.71 m/s squared. What would a rock’s velocity be 3 s after you dropped it on Mars? Since Mars has very thin atmosphere, there is no significant air resistance. The only force acting on the object is the x0 = 0m Welcome gravitational force of the planet Mars. v0 = 0 m/s to Mars Therefore the net force exerted on the rock rock is equal to the gravitational force. In gy = -3.71 m/s2 three seconds the rock will move far (toward the center of the planet), therefore we can assume that the net (gravitational) force acting on the rock is constant throughout the entire motion. Hence from Newton’s second law, at the surface of Mars, the free fall acceleration vector g is directed toward the center of Mars and its magnitude has a value of g = 3.71 m/s2. x

Solving this problem is pretty much just like the examples we have done so far for falling objects on Equations Earth. The reference frame of Mars, with the vertical Acceleration x-axis directed upward, can be used for the analysis a = -g of the motion of the rock. The rock moves with a constant acceleration. At the very moment the rock is Velocity released (t0 = 0s), its (initial) velocity will start at v = v + (a · t) 0 zero and consistently with the above figure, will be increasing in the downward direction at rate of 3.71 Position m/s per second. The vertical components of x = x + (v · t) + (½ · a · t2) 0 0 acceleration, velocity and position obey the included equations. From the equation for velocity of an object in motion with constant acceleration, the (vertical component of) velocity of the rock, at time 3s after being released is Present velocity (at 3s) = initial + acceleration * time = 0 - 3.71 m/s2 * 3 s = -11.13 m/s Three seconds after a rock is dropped on Mars its velocity will be 11.13 m/s downward (with both horizontal components of velocity being 0 m/s).

Question #4, p. 35

How far would a rock fall in 3 seconds if you dropped it on Mars? As with the previous question, we are assuming that air resistance will not be a significant factor in the calculation. This leaves only the gravitational interaction between Mars and the rock. The (magnitude of the) acceleration due to gravity (a) on Mars is 3.71m/s2 will also apply to the falling rock. The time (t) instant being investigated is 3 s after the rock has been dropped. The same reference frame of Mars can be used for the analysis of the motion of the falling rock. However, this time we will consider the position (x). Consistently with the choice of the reference frame, (the vertical components of) the rock's initial position (x0) and velocity (v0) are zero, While the (vertical component of the) acceleration is 3.71m/s2. Thus from the equation for position in a motion with constant acceleration, the (vertical component of) position of the rock, at time 3s after being released is P r e s e n t po s i t i on = i n i t i a l p o s i t i on + i n i t i a l v e l o c i t y * t i m e + ( 1 / 2 ) * a c c e l e r at i o n * ( t i m e ) 2 = = 0m + 0m / s · 0 s + ( 1/ 2 ) * ( - 3 . 7 1m / s 2 ) * 3 s  - 1 6 . 7 m Three seconds after a rock is dropped on Mars its position will 16.7m below the initial location (both horizontal components of the position of the rock have 0m value). By choosing a reference frame with the initial condition x0 = 0m, one can easily find the displacement of the rock in three second. From the initial location, the rock would fall 16.7 m in three second. The analysis of the entire motion is presented in the following table

Time

Velocity

Acceleration

Position

0s 1s

0 m/s -3.7 m/s

-3.71 m/s2 -3.71 m/s2

0m - 1.9 m

2s

-7.4 m/s

-3.71 m/s2

- 7.4 m

3s

-11.1 m/s

-3.71 m/s2

-16.7 m

Question #6 & #7, p. 35

A basketball player can leap upward .5 m. What is his initial velocity at the start of the leap? How long does the basketball player in Problem 6 remain in the air?

time

x 1.0 m

acceleration

velocity

position

0

-9.8

0

0.5m

?

-9.8

?

0

0.5m 0.5m

Ground

When the basketball player leaps off the ground, he leaves the ground with a large upward velocity (large vertical component of the velocity). As soon as the basketball player leaves the ground, only the gravitational force is exerted by the earth on the player. Therefore, he has a downward acceleration of 9.8 m/s2. Consistently with the above figure, the vertical component of the acceleration will have value of -9.8 m/s2. His initial upward velocity (vertical component of the velocity) will gradually decrease until it reaches 0. He then will begin to fall and his velocity will begin to increase in that direction (the vertical component will get more and more negative) until he returns to the ground. In a motion with constant acceleration, the position of the object is a quadratic function of time, determined by the initial position, initial velocity and the acceleration. The solution is simplest if we choose the instant when the player is at the highest point as the reference for time t0 = 0s. Consistently with all assumptions, the above table lists the information give in the problem. “present” height = “initial” height + “initial” velocity + ½ * acceleration * time2 Hence 0 m = 0.5 m + 0 m/s * t + ½ * (-9.8 m/s2) * t2 Solving for the unknown time (instant) t 4.9 m/s2 * t2 = 0.5 m t2 = 0.5 m / (4.9 m/s2)  0.1 s2

Take the square root of both sides and you get two solutions for time when the player is on the ground t1 = - 0.32 s

and

t2 = 0.32 s

The negative value of time corresponds to the instant when the player started the leap (he left the ground 0.32 s before he reached the highest point) and the positive identifies the instant when the player finished the leap (he touched the ground 0.32 s after he reached the highest point). In a motion with constant acceleration, the velocity of the object is a linear function of time, determined by the initial velocity and the acceleration “present” velocity = “initial” velocity + acceleration * time Hence at the time the player left the ground his upward velocity (vertical component) was v = 0 m/s + (-9.8 m/s2) * (-0.32s) = 3.14m/s At the start of the leap, the initial upward velocity (vertical component of velocity) of the player is 3.14 m/s. We have found that the player begins his leap 0.32 s before he reaches the highest point and ends the leap 0.32 s before he reaches the highest point. (The way up is the same amount of time as the way down.) Hence the total time he stays in air is t = 0.32 s + 0.32 s = 0.64 s One could also find that amount of time (length of the time interval) by subtracting the time (instant) of the beginning of the leap (-0.32s) from the time (instant) of the end of the leap (0.32 s) t = 0.32 s – (-0.32 s) = 0.64 s The basketball player stays in air for 0.64 s. If he jumps 0.5 m up, he does not even stay in the air for a whole second.