Projectile problem

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
My answer key says 62.5....doesnt make sense to me
Hah, yes, thats because I'm newb. The time in the air is not 5 seconds.
vertical velocity is not 25 m/s but sin(45)*25 = 17.677 m/s
So time in air is ((17.677*2)/10)=3.534 sec
So Distance is 3.534 s*17.677 m/s =62.50 meters.

Sorry about that. 😕
 
  • Like
Reactions: csx
Hah, yes, thats because I'm newb. The time in the air is not 5 seconds.
vertical velocity is not 25 m/s but sin(45)*25 = 17.677 m/s
So time in air is ((17.677*2)/10)=3.534 sec
So Distance is 3.534 s*17.677 m/s =62.50 meters.

Sorry about that. 😕
Thanks, its all good! Why do you multiply that by 2?
 
The total time in the air is the time up and the time down. So it takes 17.67 / 10m/s^2 to reach max height, and you need to double that for total duration in air.
 
  • Like
Reactions: csx
The total time in the air is the time up and the time down. So it takes 17.67 / 10m/s^2 to reach max height, and you need to double that for total duration in air.
Gotchya. Brain farted!
 
The total time in the air is the time up and the time down. So it takes 17.67 / 10m/s^2 to reach max height, and you need to double that for total duration in air.
Got another question for you if you dont mind. If an object is launched at 45 degrees and has a time of flight of 3 seconds, what is its range?
 
Got another question for you if you dont mind. If an object is launched at 45 degrees and has a time of flight of 3 seconds, what is its range?
1.5 seconds up and 1.5 seconds down.

If time up is 1.5 sec you know that initial vertical velocity is 15m/s (with gravity 10 m/s^2)
If vertical velocity is 15 m/s and its a 45 degree launch then horizontal is also 15 m/s.
15 m/s horizontal for 3 seconds is 45 meters.
 
Top