Bill is carelessly playing with a glass beaker. Bill throws the beaker up with an initial speed of 3.0 m/s at a height of 1.0 m above the ground.

Question

a- If Bill catches the beaker at the same height that he threw it at how long is the beaker in the air?

b- If Bill doesn’t catch it but instead allows it to fall down to the ground below how long would the beaker be in the air for? Note: The beaker was still initially thrown upwards

Answer 1

Hi Ariel,

How you solve this problem depends on the level of physics you are studying. If you are in a calculus based physics course, then the bulk of the work is in deriving a function which models the height of the beaker as a function of time. To do this you begin with the fact that the acceleration due to gravity on earth is 9.8m/s^2; that is, a(t)= -9.8m/s^2 (notice the negative sign is indicating the downward direction of the acceleration). Then from calculus we know that the velocity v(t) of the beaker as a function of time is an antiderivative of a(t). Using the initial velocity of 3m/s we find that v(t)=-9.8t+3. Since the height h(t) of the beaker as a function of time is an antiderivative of v(t), and we know that the initial height is one meter, we find that h(t)= -4.9t^2+3t+1. I left out a fair bit of work here, so if you are in a calculus based course definitely derive this equation in more detail so it makes sense to you!

If you are not in a calculus based physics course, then you must know/have access to the kinematic equation h(t)=1/2a(0)*t^2+v(0)*t+h(0) where a(0),v(0), and h(0) are the initial acceleration, velocity, and height, respectively. Notice that by plugging in the correct initial values for acceleration, velocity, and height into the equation results in the height equation derived in the first paragraph.

Once h(t) is known you must then solve the correct quadratic equation for a given value of h. In part (a), what is the height that Bill catches the beaker? For part (b), what is the height of the beaker when it hits the ground? Remember that you are going to be solving for t in this part of the problem, and that quadratic equations have two solutions for a given value (so you must ensure that your answer makes sense in terms of the context of the problem!)

Hope this helps!

Garrett

Answer 2

Hi Ariel!

For this question, you want to use the kinematic equations in the y direction and solve for time. y_f= y_i + v_yi*t + (1/2)*a_y*t²

For part a:

a_y=-9.8 m/s² , y_i= 1m, y_f= 1m because he catches the beaker at the same height, v_yi=3 m/s

So the equation is: 1 = 1+3t+(1/2)*-9.8*t², 0=3t+1/2*-9.8*t², which we then factor to be

0=t(3-4.9*t), t=0 seconds (because this is the height of the initial throw), and t = .612 seconds at the time of Bill catching the beaker again. So the beaker would be in the air for 0.612 seconds before being caught.

For part b the only value that changes in the equation is y_f=0m:

a_y=-9.8 m/s² , y_i= 1m, y_f= 0m because it falls to the floor, v_yi=3 m/s so the equation is:

0 = 1+3t+(1/2)*-9.8*t², 0 = 1 + 3t-4.9t², then you can use the quadratic formula to solve for the values of t where a=-4.9, b = 3, c=1. So to solve for t:

t=-b+-√b²-4ac) / 2a

So you get t= - 0.239 (can't have time be negative in this instance), and t = 0.852 seconds. So the beaker would be in the air for 0.852 seconds before hitting the ground.

Hope this helps!