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### Course: AP®︎/College Physics 1>Unit 2

Lesson 4: Horizontally launched projectiles

# Horizontally launched projectile

How to solve for the horizontal displacement when the projectile starts with a horizontal initial velocity. We also explain common mistakes people make when doing horizontally launched projectile problems. Created by David SantoPietro.

## Want to join the conversation?

• Is acceleration due to gravity 10 m/s^2 or 9.8 m/s^2? My teacher says it is 10 but Dave says it is 9.8.
• Acceleration due to gravity actually depends on your location on the planet and how far above sea level you are, and is between 9.78 and 9.83 .This is sometimes rounded up to 10 to make assignments more simple, especially when a calculator is not available, but if you're going to continue studying physics you should remember that it's closer to 9.8 .
• Why does the time remain same even if the body covers greater distance when horizontally projected? I mean when the body is just dropped without any horizontal component, it will fall straight. But when we give a horizontal velocity to the body, it should cover a parabolic path(greater than the path covered during free fall). So the body should take a longer time to fall.
: )
• The acceleration due to gravity is the same whether the object is falling straight or moving horizontally. Since acceleration is the same, then the time each object hits the ground will be the same, assuming they both start from the same height and fall the same distance.
• David mentioned that the time it takes for vertical displacement to occur would the same as the time it takes for the horizontal displacement to happen. This much makes sense, especially if air resistance is negligible.

However, what happens in the case of a cliff jumper with a wing suit? Are the times still the same for the vertical and horizontal?
• this part of physics makes me want to give up
• Why is it hard?
(1 vote)
• If something is thrown horizontally off a cliff, what is it’s vertical acceleration? How do you know? What is its horizontal acceleration? How do you know?
• Its vertical acceleration is -9.8 m/(s^2) (the acceleration due to gravity) and a projectile (if you're neglecting air resistance) never has acceleration in the horizontal direction. So in the horizontal direction the acceleration would be 0. Hope this helps!
• If in a horizontally launched projectile problem you're given the height of the 'cliff' and the horizontal distance at which the object falls into the 'water' how do you calculate the initial velocity?
• Maths version of what Teacher Mackenzie said:
Find the time it takes for an object to fall from the given height.
∆y = v_0 t + (1/2)at^2; v_0 = 0; ∆y = -h; and a = g the initial vertical velocity is zero, because we specified that the projectile is launched horizontally.
-h = (1/2)gt^2
-2h/g = t^2
√(-2h/g) = t The negative sign under the radical is fine because gravitational acceleration is also in the negative direction.
Then we take this t and plug it into the x equations
∆x = v_0t + 1/2at^2; horizontal acceleration is zero. We could also use an equation with final velocity instead of acceleration, using the understanding that final velocity will equal initial velocity.
∆x = v_0*t; solve for initial velocity
∆x/t = v_0
• Would air resistance shorten the horizontal distance you are jumping, or lengthen it?
• Air resistance would decrease your velocity, so it would shorten your horizontal jump distance.
• How would you then find the velocity when it hits the ground and the length of the hypotenuse line?
• If you were asked to find final velocity, you would need both the vertical and horizontal components of final velocity. Horizontal is easy, there is no horizontal acceleration, so the final velocity is the same as initial velocity (5 m/s). To find the vertical final velocity, you would use a kinematic equation. You have vertical displacement (30 m), acceleration (9.8 m/s^2), and initial velocity (0 m/s). You could then use the time-independent formula:
Vf^2 - Vi^2 = 2 * a * d
Vf^2 - (0)^2 = 2 * (9.8) * (30)
Vf = sqrt(2 * 9.8 * 30)
Vf = 24.2...

Now that you have the final velocity components, you can set up a right triangle to solve for the combined final velocity. The components will be the legs, and the total final velocity will be the hypotenuse. By the pythagorean theorem:
Vfx^2 + Vfy^2 = Vf^2
(5)^2 + (24)^2 = Vf^2
Vf = sqrt(5^2 + 24.2^2)
Vf = 24.7
That's the magnitude of the final velocity. To find the angle, you would need to do some trig and realize that the angle from the horizontal is opposite to Vfy and adjacent to Vfx. We can write this as:
tan(theta) = Vfy / Vfx
theta = atan ((24.2) / (5))
theta = 78.3... degrees

And there you have both the magnitude and angle of the final velocity. Hope this helps! (Yes, I am the slightest bit too lazy to actually write the symbol for theta)