Find Horizontal Range of Projectile with Initial Speed of 60m/s

In summary, the projectile is launched with an initial speed of 60m/s and reaches a maximum height of 150m. To find the horizontal range, the equation v=gt can be used to solve for the total flight time, which is then doubled to get the total flight time. This gives a horizontal range of 284m.
  • #1
jegues
1,097
3

Homework Statement


A projectile is launched with an intial speed of 60m/s. If it reaches a maximum height of 150m above this level surface then its horizontal range (in m) is ____ ?


Homework Equations



v[tex]^{2}[/tex] = vo[tex]^{2}[/tex] + 2ay

y = yo + Vot + 1/2at^2

The Attempt at a Solution



Vo = 60m/s, then Vox = 60cos(theta); Voy = 60sin(theta) y-yo = 150m

V^2 = Vo^2 + 2ad

0 = 60sin(theta) + 2(-9.8)(150)

theta = 64.6 degrees, so Voy = 54.2m/s

Then,

y-yo = Voyt + 1/2at^2

0 = -150 + 54.2t - 4.9t^2

no zeroes, I was going to grab the zeros for this quadratic and the use, x/2 = Vox/t; x being my final range. Any ideas?

EDIT: I thought about it a bit more and decided that since I already found the angle I could just turn it into a triangle problem.

the triangle has a height of 150, so the distance in the x direction is simply, x = 150/tan(theta);

This will give the HALF of the horizontal range, so 2x will give me my answer. Now oddly enough, 4x seems to give me the correct answer and 2x only half of the correct answer... Coincidence?
 
Last edited:
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  • #2
If you are having trouble solving for t using your quadratic equation, why not use v=gt to solve for t, then double it to get the total flight time. The triangle approach does not work, because the projectile takes on a parabolic shape.
 
  • #3
Doh, Vfy = Voy -gt1; t1 = Voy/g ; t2 the full time, therefore t2 = 2t1

x = Vox * t2 = 284m,

Thanks.
 

Related to Find Horizontal Range of Projectile with Initial Speed of 60m/s

1. How do you calculate the horizontal range of a projectile with an initial speed of 60m/s?

The horizontal range of a projectile can be calculated using the formula R = (v2sin2θ)/g, where R is the horizontal range, v is the initial speed of the projectile, θ is the angle of launch, and g is the acceleration due to gravity (9.8m/s2).

2. What is the significance of calculating the horizontal range of a projectile?

Calculating the horizontal range of a projectile allows us to determine how far the projectile will travel horizontally before hitting the ground. This information is important in various fields such as sports, engineering, and military applications.

3. How does changing the initial speed of a projectile affect its horizontal range?

Changing the initial speed of a projectile will directly affect its horizontal range. As the initial speed increases, the horizontal range will also increase. Similarly, decreasing the initial speed will result in a shorter horizontal range.

4. Is the horizontal range of a projectile affected by the angle of launch?

Yes, the horizontal range of a projectile is affected by the angle of launch. The maximum horizontal range is achieved when the angle of launch is 45 degrees. Launching the projectile at a higher or lower angle will result in a shorter horizontal range.

5. What are some real-life examples of projectiles with an initial speed of 60m/s?

Some real-life examples of projectiles with an initial speed of 60m/s include a baseball pitch, a golf ball hit with a driver, and a bullet shot from a rifle. These examples can vary depending on factors such as air resistance and the angle of launch.

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