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### Course: Physics archive>Unit 2

Lesson 2: Optimal angle for a projectile

# Optimal angle for a projectile part 3: Horizontal distance as a function of angle (and speed)

In this video, we'll finish writing the formula for distance as a function of the angle. Created by Sal Khan.

## Want to join the conversation?

• I have two questions.
1) What does "rate" mean in this video?
2) The formula we got was d=2s^2*sin(theta)*cos(theta)/g, but when I learnt it we got it as d=s^2*sin2(theta)/g.
Can anybody explain why?
Thanks. :)
• yes..rate means velocity!!
and sin2(theta)=2sin(theta)cos(theta)
• How does x and y component specify speed of the particle ?
I mean of the object is projected at an angle of 30degrees then how does its speed will be along x and y ?
• An object thrown at an angle with an initial speed can be modeled as two separate velocities going in different directions, one vertical and one horizontal. Whether thrown at an angle with a velocity, or pushed horizontally and vertically simultaneously, it will cause the same effect on the object and will act the same way (it will have a curved path), and will be mathematically equivalent than the first method mentioned.
• Isn't sin(theta)*cos(theta)=(1/2)*sin(2*theta) by reversing that sin(2*theta)=sin(theta+theta) and then we could use the sine addition identity and get sin(2*theta)=2*sin(theta)*cos(theta).
We have this in the distance formula so could it be written as (s^2/g)*sin(2*theta)?
• Yes, it can indeed. Many textbooks phrase the formula like that instead, probably because it's a little easier to memorize.
(1 vote)
• why is rate s(cos theta) and not just s?
• s(cos theta) is the horizontal component of the s vector. The s vector by itself has a vertical component as well. Since s has a vertical component, it will not give us the correct value for horizontal distance. Hope that helps anyone with the same question.
• How would the ending equation end if the initial height wasn't 0?
• In that case, you'd have a different value for time, and therefore a different value for Δx.
(1 vote)
• Totally different question but - would changing the mass of a spherical projectile affect the horizontal displacement when launched horizontally from a constant height above the ground, with the same initial velocity in the presence of air? For example, say a lighter (8 grams) and heavier (34 grams) projectile were put to test.
• Of course it would. The reason being that a lighter object has less inertia compared to a heavier object. So, while the lighter object will slow down quicker, the heavier one will go farther due to its mass. In other words, the heavier object overcomes more of the air resistance (Though we don't consider air resistance in problems, we can't explain many stuff without air resistance) compared to the lighter object.

Hope this helped!
(1 vote)
• at sal says that we have to find horizontal distance. why can't we use vertical distance?
(1 vote)
• Because "optimal angle" means the one that will make it go the furthest, not the highest.
• Horizontal distance is the range right?
(1 vote)
• yes, range is the maximum horizontal distance traveled by a projectile.
(1 vote)
• Is horizontal distance the same as displacement?
(1 vote)
• Well...Displacement is the distance an object traveled but with a magnitude and direction. If an object's 'a' point(starting point) and 'b' point(ending point) are in a straight horizontal line, then the displacement is the horizontal distance but with both magnitude and direction. But if points a and b are not aligned horizontally, such as if point a was on the bottom of a hill and b was on the top, then, the displacement is horizontal distance plus vertical distance with both magnitude and direction. Tell me if I wrote this confusing. Or someone out there can give a better explanation.
(1 vote)
• how do I calculate the time an object is in the air given the velocity and the angle at which it was launched
(1 vote)
• First you find the initial velocity's vertical component which is velocity*sin(angle). Then use that initial upwards velocity along with delta y=0 and the acceleration (-9.8 m/s^2 on earth) to calculate the time it spends in the air.

Did that help?