QuestionA) The horizontal velocity increases over time. (B) The horizontal velocity remains constant throughout the projectile's flight. C) The horizontal distance traveled increases linearly with time squared.
2. What is the effect of gravity on a projectile's horizontal motion? A) Gravity pulls the projectile downward but has no effect on horizontal motion. B) Gravity accelerates the projectile horizontally. C) Gravity causes the horizontal velocity to decrease.
3. When a projectile is launched at an angle of $45^{\circ}$ above the horizontal, what happens

Studdy Solution

STEP 1

What is this asking? This is asking about how gravity affects the horizontal motion of a projectile, and how the horizontal distance changes over time.
It also asks about the special case of launching at a 45-degree angle. Watch out! Don't mix up horizontal and vertical motion!
Gravity only affects vertical motion.

STEP 2

1. Horizontal Velocity
2. Gravity's Effect
3. 45-Degree Launch

STEP 3

Alright, imagine throwing a baseball perfectly horizontally.
Ignoring air resistance, is there anything pushing or pulling it sideways?
Nope!

STEP 4

That means there's no horizontal acceleration!
If there's no acceleration, the horizontal velocity stays constant!

STEP 5

Gravity pulls everything downwards.
It's like an invisible string pulling straight down towards the ground.

STEP 6

Does that string pull sideways at all?
No way!
So, gravity affects the vertical motion, making the projectile fall, but it doesn't touch the horizontal motion.

STEP 7

Launching at 45 degrees is special!
It gives you the maximum horizontal range if you're landing at the same height you launched from.

STEP 8

Think of it like this: If you launch straight up, you get zero horizontal distance.
If you launch horizontally, you get some distance.
45 degrees is the perfect balance between going up and going forward.

STEP 9

Let's say you launch a projectile with an initial velocity $v_0$ at an angle $\theta$ .
The horizontal velocity is given by $v_x = v_0 \cdot \cos(\theta)$ .
When $\theta = 45^\circ$ , we have $v_x = v_0 \cdot \cos(45^\circ) = v_0 \cdot \frac{\sqrt{2}}{2}$ .

STEP 10

The horizontal distance $x$ traveled over time $t$ is given by $x = v_x \cdot t$ .
Since $v_x$ is constant, the horizontal distance increases linearly with time, not with time squared!

STEP 11

1. (B) The horizontal velocity remains constant throughout the projectile's flight.
2. (A) Gravity pulls the projectile downward but has no effect on horizontal motion.
3. When a projectile is launched at an angle of 45 degrees above the horizontal, it achieves the maximum horizontal range for a given initial velocity (assuming it lands at the same height it was launched from).

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Studdy Solution

STEP 1

STEP 2

1. Horizontal Velocity
2. Gravity's Effect
3. 45-Degree Launch

STEP 3

Alright, imagine throwing a baseball perfectly horizontally.
Ignoring air resistance, is there anything pushing or pulling it sideways?
Nope!

STEP 4

That means there's no horizontal acceleration!
If there's no acceleration, the horizontal velocity stays constant!

STEP 5

Gravity pulls everything downwards.
It's like an invisible string pulling straight down towards the ground.

STEP 6

Does that string pull sideways at all?
No way!
So, gravity affects the vertical motion, making the projectile fall, but it doesn't touch the horizontal motion.

STEP 7

Launching at 45 degrees is special!
It gives you the maximum horizontal range if you're landing at the same height you launched from.

STEP 8

Think of it like this: If you launch straight up, you get zero horizontal distance.
If you launch horizontally, you get some distance.
45 degrees is the perfect balance between going up and going forward.

STEP 9

Let's say you launch a projectile with an initial velocity $v_0$ at an angle $\theta$ .
The horizontal velocity is given by $v_x = v_0 \cdot \cos(\theta)$ .
When $\theta = 45^\circ$ , we have $v_x = v_0 \cdot \cos(45^\circ) = v_0 \cdot \frac{\sqrt{2}}{2}$ .

STEP 10

The horizontal distance $x$ traveled over time $t$ is given by $x = v_x \cdot t$ .
Since $v_x$ is constant, the horizontal distance increases linearly with time, not with time squared!

STEP 11

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Studdy Solution

STEP 1

STEP 2

1. Horizontal Velocity2. Gravity's Effect3. 45-Degree Launch

STEP 3

Alright, imagine throwing a baseball perfectly horizontally.Ignoring air resistance, is there anything pushing or pulling it sideways?Nope!

STEP 4

That means there's no horizontal acceleration!If there's no acceleration, the horizontal velocity stays **constant**!

STEP 5

Gravity pulls everything *downwards*.It's like an invisible string pulling straight down towards the ground.

STEP 6

Does that string pull sideways *at all*?No way!So, gravity affects the vertical motion, making the projectile fall, but it doesn't touch the horizontal motion.

STEP 7

Launching at 45 degrees is special!It gives you the **maximum horizontal range** if you're landing at the same height you launched from.

STEP 8

Think of it like this: If you launch straight up, you get zero horizontal distance.If you launch horizontally, you get *some* distance.45 degrees is the perfect balance between going up and going forward.

STEP 9

STEP 10

The horizontal distance xxx traveled over time ttt is given by x=vx⋅tx = v_x \cdot tx=vx​⋅t.Since vxv_xvx​ is constant, the horizontal distance increases **linearly** with time, *not* with time squared!

STEP 11

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Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Horizontal Velocity
2. Gravity's Effect
3. 45-Degree Launch

Alright, imagine throwing a baseball perfectly horizontally.
Ignoring air resistance, is there anything pushing or pulling it sideways?
Nope!

That means there's no horizontal acceleration!
If there's no acceleration, the horizontal velocity stays constant!

Gravity pulls everything downwards.
It's like an invisible string pulling straight down towards the ground.

Does that string pull sideways at all?
No way!
So, gravity affects the vertical motion, making the projectile fall, but it doesn't touch the horizontal motion.

Launching at 45 degrees is special!
It gives you the maximum horizontal range if you're landing at the same height you launched from.

Think of it like this: If you launch straight up, you get zero horizontal distance.
If you launch horizontally, you get some distance.
45 degrees is the perfect balance between going up and going forward.

The horizontal distance $x$ traveled over time $t$ is given by $x = v_x \cdot t$ .
Since $v_x$ is constant, the horizontal distance increases linearly with time, not with time squared!