Projectile Motion Problem: Calculating Time of Flight and Distance Traveled"

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A projectile is launched at a 20-degree angle with an initial velocity of 15 m/s, prompting a discussion on calculating its time of flight and horizontal distance. The original calculations yielded a vertical distance of 11.48 m, horizontal distance of 10.13 m, and a total time of 3.06 seconds, but these were questioned for accuracy. Participants suggested using standard projectile motion equations to derive the correct values, emphasizing the importance of understanding the equations for accurate results. A hint was provided to set the vertical position equation to zero to find the time of flight. The conversation highlighted the need for clarity in applying physics concepts rather than focusing on grammar.
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Before you get all worked up, I am just going to let you know that this isn't homework. Here is my problem : A projectile is fired at a 20 degree angle with a velocity of 15 m/s. What is the total time of flight? What is the distance of its hight? How far did it travel in the horizontal direction?

Ok, no to solve this i used Vf^2=Vi^2-2ad, (this is on the trip up) where Vf = 0, Vi = 15, and a= 9.8, to give me the vertical distance of 11.48m. To find the horizontal distance, i used 15cos20 as the intial velocity and plugged it into the same equation to get a hor. distance of 10.13m. Then, to find total time i used a= (Vf-Vi)/t, where on the trip up a=9.8, Vf =0, and Vi = 15 to get a time of 1.53 sec, on the way down i used a=9.8, Vf=15, and Vi = 0 in the same equation to get the same exact time, which combine gave me a total of 3.06s...so all of my answers are as follows:
Dv=11.48m, Dh=10.13m, Ttotal=3.06s
If anyone could confirm or correct my process id be greatly appreciated. Also I am new so if this is indeed the wrong board i would gladly be redirected:smile: Thanks in advance
 
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Your answers don't seem correct.

Use the 'standard' equations for projectile motion:

v_{x}(t) = v_{0}\cos(\alpha)
v_{y}(t) = v_{0}\sin(\alpha) - gt
x(t) = v_{0}\cos(\alpha)t
y(t) = v_{0}\sin(\alpha)t - \frac{1}{2}gt^2.
 
Wow, my teachers hasnt given us those equations so I am really not sure how to use them








radou said:
Your answers don't seem correct.

Use the 'standard' equations for projectile motion:

v_{x}(t) = v_{0}\cos(\alpha)
v_{y}(t) = v_{0}\sin(\alpha) - gt
x(t) = v_{0}\cos(\alpha)t
y(t) = v_{0}\sin(\alpha)t - \frac{1}{2}gt^2.
 
! said:
Wow, my teachers hasnt given us those equations so I am really not sure how to use them

How to use them? Simply read them. :wink:

(Hint: use the condition y(t) = 0 to get the time of the flight t. Then you can find all you need.)
 
Wow! Excuse my terrible english in the previous statement..."my teachers hasnt...", you must think I am retarded
radou said:
How to use them? Simply read them. :wink:

(Hint: use the condition y(t) = 0 to get the time of the flight t. Then you can find all you need.)
 
! said:
Wow! Excuse my terrible english in the previous statement..."my teachers hasnt...", you must think I am retarded

Don't worry, this is no grammar forum. :wink:
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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