random projectile motion question

[cloak25]

Was just reviewing some projectile motion stuff and stumbled upon this question from TBR:

What is the range for a projectile launched at a 45 degree angle that has a flight time of 3 seconds?

answer is 44.1m

TBR explanation:
with a flight time of 3s, it takes 1.5s to reach its apex. A 1-second drop time covers 5m and a 2-second drop time covers 20m, so a 1.5 sec drop covers about 11-12m, so the maximum height is about 11-12m. The range for a 45 degree launch is four times the maximum height, so R = 44-48m.

why can't I use t = 2Vo/g to find Vo and then use R=Vo^2(sin2theta)/g to find the range? (I found these equations from the chapter).

according to TBR, the equation above (t=2Vo/g) represents total time in air. So.. what am I doing wrong?: 3 = 2Vo/g. solve for Vo = 15. R = Vo^2(sin90)/g = 22.5m 😕

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[rjosh33]

Was just reviewing some projectile motion stuff and stumbled upon this question from TBR:

What is the range for a projectile launched at a 45 degree angle that has a flight time of 3 seconds?

answer is 44.1m

TBR explanation:
with a flight time of 3s, it takes 1.5s to reach its apex. A 1-second drop time covers 5m and a 2-second drop time covers 20m, so a 1.5 sec drop covers about 11-12m, so the maximum height is about 11-12m. The range for a 45 degree launch is four times the maximum height, so R = 44-48m.

why can't I use t = 2Vo/g to find Vo and then use R=Vo^2(sin2theta)/g to find the range? (I found these equations from the chapter).

according to TBR, the equation above (t=2Vo/g) represents total time in air. So.. what am I doing wrong?: 3 = 2Vo/g. solve for Vo = 15. R = Vo^2(sin90)/g = 22.5m 😕

I'm with you as far as solving for Voy. I used a different approach, but got 14.7 m/s (basically 15) as well.

Now, ignoring air resistance, R simply equals Vox times time (this is a key equation used to help solve many a projectile problem). Since the launch was at a 45 degree angle, Voy = Vox.

Vox = Voy = 14.7 m/s. Time was given to us as t = 3s.

So, R = (Vox)(t) = (14.7)(3) = 44.1 m

Make sense?

 

[MedPR]

Was just reviewing some projectile motion stuff and stumbled upon this question from TBR:

What is the range for a projectile launched at a 45 degree angle that has a flight time of 3 seconds?

answer is 44.1m

TBR explanation:
with a flight time of 3s, it takes 1.5s to reach its apex. A 1-second drop time covers 5m and a 2-second drop time covers 20m, so a 1.5 sec drop covers about 11-12m, so the maximum height is about 11-12m. The range for a 45 degree launch is four times the maximum height, so R = 44-48m.

why can't I use t = 2Vo/g to find Vo and then use R=Vo^2(sin2theta)/g to find the range? (I found these equations from the chapter).

according to TBR, the equation above (t=2Vo/g) represents total time in air. So.. what am I doing wrong?: 3 = 2Vo/g. solve for Vo = 15. R = Vo^2(sin90)/g = 22.5m 😕

Because the equation is actually t=2voy/g, not 2vo/g. If voy is 15, vo = 15/sin45.

15/sin45 = ~20. 20^2 = 400. 400*sin90=400. 400/10 = 40.

Can someone tell me why the following doesn't work?

You use t=1.5 to find out that the height of the projectile is ~11m. If you make a triangle where y=11 and theta is 45deg, you find that x is also 11. Since 2x = the range, shouldn't the range be 22?

I can see from using kinematics equations that it is ~40, but why doesn't the triangle work?