A ball is thrown down vertically with an initial speed of v0 from a height of h. (a) What is its speed just before it strikes the ground? (b) How long does the ball take to reach the ground? What would be the answers to (c) part a and (d) part b if the ball were thrown upward from the same height and with the same initial speed? Before solving any equations, decide whether the answers to (c) and (d) should be greater than, less than, or the same as in (a) and (b).

Answer:

Let us resolve the problem into two parts as indicated by your question, starting with when the ball is thrown downwards.

(a) The speed of the ball just before it strikes the ground when thrown downwards with an initial speed \( v_0 \) can be determined using the formula derived from the kinematic equation for uniformly accelerated motion:

\[ v = \sqrt{v_0^2 + 2gh} \]

Here:

- \( v \) is the final speed - \( v_0 \) is the initial speed - \( g \) is the acceleration due to gravity (approximately \( 9.8 m/s^2 \) on Earth) - \( h \) is the initial height from which the ball is thrown

(b) The time it takes for the ball to reach the ground can be found using the following equation, which also comes from the kinematic equations:

\[ v = v_0 + gt \] \[ t = \frac{v - v_0}{g} \]

Since we calculated \( v \) in part (a), we can use it here to solve for \( t \).

Now let's consider when the ball is thrown upwards.

(c) If the ball were thrown upwards with the same initial speed \( v_0 \), its speed just before it strikes the ground would actually be greater than in part (a). This is because the ball has to first decelerate going up until its velocity becomes 0 at its peak, and then it will accelerate due to gravity on its way back down, effectively having more time to gain speed. The equation to find the final speed would still be the same, but you would account for the ball going up and then coming down.

(d) Concerning the time it would take the ball to reach the ground when thrown upwards, it would take a longer time than when thrown downward with the same initial speed \( v_0 \). This is because the ball would spend extra time reaching the highest point before falling back down to the ground.

Now let's compare the answers to questions (a) and (b) with (c) and (d):

(c) should be 'greater than' (a) (d) should be 'greater than' (b)

To be clearer: for part (c), the speed is greater because the ball has more distance to accelerate due to gravity when thrown upwards. For part (d), the time is greater since it includes the time for the ball going upward and then the time for it to fall back down.

Extra:

Understanding the basic principles at work here is key for students to grasp the concepts involved in projectile motion and free fall.

- When an object is thrown vertically downwards, it's already moving in the direction of the gravitational pull, so its speed increases due to both its initial velocity and the acceleration due to gravity. - When you throw an object upwards, the initial motion is against gravity, so the object slows down until it stops momentarily at its highest point. After that point, it's accelerating downwards just like it was dropped from rest at that height. So it's effectively like a two-stage motion: upwards against gravity, and then downwards with gravity. - The two main kinematic equations used here can be derived from the basic equations of motion assuming constant acceleration. They relate speed, acceleration, time, and distance and are very powerful tools for analyzing motion in one dimension.

Understanding the equations of motion and how to apply them will really help students to solve a variety of physics problems related to motion, not only in the vertical direction as in this scenario but also horizontally and at angles, which leads into two-dimensional projectile motion.