what are the steps to solving these questions? i need help with ballistics

Question

Ballistics problem Frodo Baggins is about to throw the One Ring into Mount Doom . He is standing at the edge of the crater , which is 15 meters wide and full to the rim with lava . Frodo throws the ring from ground level with a vertical speed of 9.5 meters per second , and gravity is 9.8 meters per second squared . 

a ) If Frodo throws the ring straight up , how long will it take to fall into the lava ? 

b) Gollum , who is hiding behind Frodo , will try to catch the ring when it is at its highest point How high does Gollum need to jump , and how many seconds after Frodo throws the ring will it be at its highest point ?

c ) Frodo is a Hobbit , so it is a good approximation to say that he throws the ring from ground level . If Frodo was an Ent , he would throw the ring from 8 meters above ground level . How long would the ring take to land in the lava if he threw it upward ? 

d ) If Frodo wants to give the ring horizontal velocity as well as his usual vertical velocity , so it will land in the lava at the opposite side of the crater , what horizontal velocity should he give it ?

Answer 1

In long physics problems like these, it's usually beneficial to draw a picture if one isn't given to you. Here's one I drew: https://ibb.co/85jvGYp. Just something small, enough to organize your thinking about what's going on. Let's say for this problem, up is positive and down is negative. We are given the following quantities in the problem:

v_y = 9.5 m/s

a_y = -9.8 m/s² (downwards)

∆x = 15m

∆y = 0m (because the crater is filled to the brim with lava)

A) Here, Frodo is throwing the ring straight up and we want to know how long it will take to fall. Well, we know that this is just a 1-dimensional problem, and so easier because of that. All we have to do is think about which kinematic equation allows us to incorporate what we do know (initial velocity and acceleration) and time. Well, that sounds like ∆y = v_0yt + 1/2 a_y t²:

We know ∆y is 0, so this allows us to say -v_0yt = 1/2 a_y t²

Taking out a t from both sides, we get -v_0y = 1/2 a_y t

And finally solving for t: t = -2v_0y / a_y

t = -2 (9.5 / -9.8) = 1.938 sec

B) Now Gollum wants to catch the ring when it's at its highest point. This would be halfway through its motion, since we don't have any change in height. So you can just say the time for this is the total time cut in half, or about .969 sec. Of course, if you're unsure you can always check yourself by using the kinematic equations to find the time when the velocity becomes 0 (meaning the projectile is turning). To find the height, we can just use the same equation as in part A), just with our time already plugged in:

∆y = v_0yt + 1/2 a_y t²

∆y = 9.5(.969) + 1/2 (-9.8)(.969²)

∆y = 4.605 m

C) Now this part says to do this same calculation, but if we were 8 feet off the ground. This means our ∆y will be -8m (signs are important!). We just keep doing the same thing pretty much:

∆y = v_0yt + 1/2 a_y t²

However, this time solving for t is not as easy, because we can't just factor one t out, making the equation linear. Instead, we have to do a quadratic formula. (Or, if you have a calculator and don't have to show your work for the algebra, you can just plug things in from the get go. But it's usually good form to solve for the variable of interest algebraically first and then plug things in.)

∆y = v_0yt + 1/2 a_y t²

We can put ∆y on the same side as everything else and make this into a quadratic with the coefficient of t² being a, the coefficient of t being b, and the constant being c:

1/2 a_y t² + v_0yt - ∆y = 0

t = (-b +/- √(b² - 4ac) ) / 2a

t = (-v_0y +/- √ (v_0y² - 4(1/2 a_y)(∆y) ) ) / 2(1/2 a_y)

t = (-9.5 +/- √ (9.5² - 4(1/2)(-9.8)(-(-8)) ) ) / 2 (1/2(-9.8))

t = 2.57 sec (taking only the positive solution because negative time doesn't make sense)

D) Now we want the ring to touch the ground (lava) at the far end of the crater, so 15m across. Here, we finally have 2 dimensions to consider, so we need to think about 2 equations. We know that there is no acceleration in the x-direction, so we can use that. We also know that this time Frodo is throwing it, so ∆y is back to being 0 (yay). In fact, we don't even need to set up 2 equations because one of them we've already done:

∆y = v_0yt + 1/2 a_y t²

t = -2v_0y / a_y = 1.938 sec, regardless of the x-velocity

∆x = v_0xt

v_0x = ∆x / t

v_0x = 15 / 1.938 = 7.74 m/s

Hope this helps! If you have any questions about the process for these or similar problems make sure to leave a comment!