Physics from BR Question

[whoknows87]

Ok this might seem silly but I'm not sure if I misread something or I'm totally blanking out example 1.14 asks what is the range for a projectile launched at 45 degrees with a flight time for 3 sec, the answer goes and it takes 1.5 sec to reach the apex t=Vo/a understood that so 1 second covers 5m ( honestly not sure where the 5 m came from anyways so 1 sec = 5 , 2 sec = 20 so 1.5 sec = 11-12 m the range for 45 is ( 4times the maxim height?

on the previous page it says the max height is h ( only with a 90) max range is 2H with ( 45)

soooo why isnt the range for the 45 in the question 2 times ( 11-12) where did the 4 come from am I missing something real simple here?

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[chiddler]

i'm having difficulty understanding your question. could you please write it a little bit more clearly

 

[whoknows87]

Sorry I tried to copy the question from TBR book and their explanation , lets try this again the question as written in the book * What is the range for a projectile launched at 45 that has a flight time of 3 seconds?

** the provious page addressed this range and height issues and this is exactly what it said ' the max height that a projectile can possibly obtain is H ( possible only with a 90 angle, while the max range is 2H ( possible only with 45)

so the range for a projectile is 2h or 4h? the answer key says a 1.5 second drop cover about 11-12m so the max is about 11-12, the range for a 45 is four times the max height

I thought the range was 2H for a 45 what am I missing or misreading here?

 

[whoknows87]

Bump any help is appreciated...

 

[shoehornlettuce]

Ok this might seem silly but I'm not sure if I misread something or I'm totally blanking out example 1.14 asks what is the range for a projectile launched at 45 degrees with a flight time for 3 sec, the answer goes and it takes 1.5 sec to reach the apex t=Vo/a understood that so 1 second covers 5m ( honestly not sure where the 5 m came from anyways so 1 sec = 5 , 2 sec = 20 so 1.5 sec = 11-12 m the range for 45 is ( 4times the maxim height?

on the previous page it says the max height is h ( only with a 90) max range is 2H with ( 45)

soooo why isnt the range for the 45 in the question 2 times ( 11-12) where did the 4 come from am I missing something real simple here?

Think of it like this.

Since you are launching at a 45 degree angle both components of velocity are going to be equal (VoX = VoY).

VoX = initial X component of velocity
VoY = initial Y component of velocity

You know the object is in the air for 3 seconds so it must take 1.5 seconds for it to reach its maximum height. Remember this equation: (Vy = Voy + ayt) where:

Vy = velocity at time t
Voy = initial velocity
ayt = acceleration

Now solve the equation for the initial Y component of velocity using 1.5 seconds. You know at 1.5 seconds Vy must be 0 because the object is at its maximum height. Acceleration is g which is 9.8 m/s so:

Vy = Voy + ayt
0 = Voy + (9.8)(1.5)
so Voy = 14.7 m/s

This gives you your initial y component of velocity. The angle is 45 degrees so the y component must equal the x component. So the initial x component of velocity is also 14.7 m/s. Now you simply multiply 14.7 * 3sec (the total time in flight) and you should get 44.1m.

 

[MAnne]

I'm confused on this too. On page 30 (2010) it says the maximum range is 2h launched at 45 degrees. In Example 1.14 explanation, it says that the maximum height is 4 time the height.

 

[chiddler]

ok lets see. lets use their numbers. They tell us that it is 1.5 seconds from bottom to top, and the same from top to bottom. This is also 11.25 meters. so the average velocity should be, for one way, 15 m/s. This is the Y speed.

since it is 45 degrees, x speed and y speed are equivalent. that means 3 seconds, the projectile moves at 15 meters per second to the right. 15*3 = 45 meters range.

ok lets find maximum height at 90 degrees upwards launch.

magnitude of initial speed is 15^2 + 15^2 = x^2.

x = 21.2 m/s magnitude of initial speed.

lets see how high it goes when at 90 degrees up.

0.5 mv^2 = mgh
0.5 v^2 = gh
0.5 v^2 / g = h
h = 0.5 * (21.2)^2 / 10 = 22.4!

MAX height is 22.4.

what about height when it is 45 degrees? as we said earlier, 11.25.

So. Here is our algorithm:

Max range, at 45 degrees, is R. Max height, at 90 degrees, is H.

When at max range, height is H/2.
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i typed this up a while ago in a pm exchange so i don't remember details. good luck.