Projectile Angle and Distance problem

In summary, the projectile was launched upward from the top of a 100m tower with an initial velocity of 36m/s. It reached a maximum height of 66m after 8 seconds.
  • #1
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Homework Statement



A projectile was launched from the top of a 100m tower upward at an angle theta to the horizontal with an initial velocity of 36m/s. The time of flight of the projectile was 8 seconds.

Determine:
i) The angle of projection
ii) The maximum height reached by the projectile

Homework Equations



S=ut+1/2at^2

The Attempt at a Solution



I am used to solving these questions when i am not given time but given both the horizontal and vertical distances. At which I use the quadratic formula to solve for theta. However in this case I can not do that as when i split my horizontal and vertical components and apply the above equations I am left with these.

Horizontally
S=36cos(?)*8

Vertically
-100=36sin(?)*8+0.5*-9.81*64

I have tried rearranging the vertical formulas to solve for theta

-1/2*-9.81*64-100/8=36sin(?)
301.42 = 36*sin(?)
sin-1(36/301.42) = 6.85 degrees

I really don't think that the answer is correct. The lecture who set the question will not give any answers.
 
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  • #2
Nate-2016 said:
Vertically
-100=36sin(?)*8+0.5*-9.81*64
OK

I have tried rearranging the vertical formulas to solve for theta

-1/2*-9.81*64-100/8=36sin(?)
You have only the 100 being divided by 8. But is that correct?
 
  • #3
Yeah. Only the 100 divided by 8. I am thinking that maybe 6.85 degrees is the correct angle for a projectile to fall from 100 metres in 8 seconds.
 
  • #4
Nate-2016 said:
Only the 100 divided by 8.
Check the algebra on this. I don't think it's correct to have just the 100 divided by 8.
 
  • #5
Im not exactly sure on the rerrangment to solve for U. As i thought my 1st rearrange was correct.

I tried the rearrange dividing the whole thing by 8 = sin-1(26.74/36) = 47.96 degrees.

I tried the rearrange dividing by 16 from 1/2*8 under everything = sin-1(36/60.76) = 36.33 degrees
 
  • #6
s = ut + 1/2at^2
ut = s - 1/2at^2
u = (s - 1/2at^2)/t
36sin(?) = (-100-1/2*-9.81*64)/8

Sin(?)= 26.74/36
Sin-1(26.74/36) = (?)
 
  • #7
Nate-2016 said:
s = ut + 1/2at^2
ut = s - 1/2at^2
u = (s - 1/2at^2)/t
36sin(?) = (-100-1/2*-9.81*64)/8

Sin(?)= 26.74/36
Sin-1(26.74/36) = (?)
That looks correct.
 

1. What is the basic concept of a projectile angle and distance problem?

The basic concept of a projectile angle and distance problem is to determine the angle and velocity at which a projectile must be launched in order to hit a specific target at a given distance.

2. How is the angle of projection related to the distance travelled by a projectile?

The angle of projection directly affects the distance travelled by a projectile. The greater the angle, the further the projectile will travel, as long as the initial velocity remains constant.

3. What factors affect the distance travelled by a projectile?

The distance travelled by a projectile is affected by the initial velocity, angle of projection, air resistance, and gravity. Other factors such as wind and elevation can also impact the distance travelled.

4. How do you calculate the maximum height reached by a projectile?

The maximum height reached by a projectile can be calculated using the formula h = (v^2 * sin^2 theta) / 2g, where h is the height, v is the initial velocity, theta is the angle of projection, and g is the acceleration due to gravity.

5. What are some real-life applications of projectile angle and distance problems?

Projectile angle and distance problems are commonly used in fields such as engineering, ballistics, and sports. They can be applied to situations such as designing bridges and buildings, calculating trajectories for missiles and projectiles, and determining the optimal angle for a golf shot or a baseball pitch.

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