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Math  /  Geometry

Question4) A person throws a ball with an initial velocity of 15 meters /sec/ \mathrm{sec} at an angle of 20∘20^{\circ} above the grd How far from the person will the ball land? Horizo

Studdy Solution

STEP 1

What is this asking? How far does a ball travel horizontally if thrown at 15 m/s15 \text{ m/s} at a 20∘20^\circ angle? Watch out! Don't forget to break the initial velocity into horizontal and vertical components!

STEP 2

1. Find the horizontal and vertical components of the initial velocity.
2. Calculate the time the ball is in the air.
3. Calculate the horizontal distance.

STEP 3

Alright, let's **break down** that initial velocity!
We've got 15 m/s15 \text{ m/s} at a 20∘20^\circ angle.
We need to find the horizontal and vertical parts of this velocity.
Think of it like a right triangle, where the initial velocity is the hypotenuse!

STEP 4

The horizontal component, let's call it vxv_x, can be found using cosine: vx=15 m/s⋅cos⁡(20∘)≈15 m/s⋅0.94≈14.1 m/sv_x = 15 \text{ m/s} \cdot \cos(20^\circ) \approx 15 \text{ m/s} \cdot 0.94 \approx \mathbf{14.1 \text{ m/s}} So, the ball is moving horizontally at roughly 14.1 m/s\mathbf{14.1} \text{ m/s}.

STEP 5

Now, for the vertical component, vyv_y, we'll use sine: vy=15 m/s⋅sin⁡(20∘)≈15 m/s⋅0.34≈5.1 m/sv_y = 15 \text{ m/s} \cdot \sin(20^\circ) \approx 15 \text{ m/s} \cdot 0.34 \approx \mathbf{5.1 \text{ m/s}} That's the **initial vertical speed**, about 5.1 m/s\mathbf{5.1} \text{ m/s} upwards!

STEP 6

To find the **hang time**, we'll focus on the vertical motion.
We know the initial vertical velocity (vyv_y) and the acceleration due to gravity (approximately g=−9.8 m/s2g = -9.8 \text{ m/s}^2, negative since it's downwards).
At the highest point, the vertical velocity is zero.

STEP 7

We can use this equation: vf=vi+atv_f = v_i + at, where vfv_f is the final vertical velocity (zero at the highest point), viv_i is the initial vertical velocity (5.1 m/s5.1 \text{ m/s}), aa is the acceleration (−9.8 m/s2-9.8 \text{ m/s}^2), and tt is the time to reach the highest point. 0=5.1 m/s+(−9.8 m/s2)⋅t0 = 5.1 \text{ m/s} + (-9.8 \text{ m/s}^2) \cdot t

STEP 8

Solving for tt: t=−5.1 m/s−9.8 m/s2≈0.52 st = \frac{-5.1 \text{ m/s}}{-9.8 \text{ m/s}^2} \approx \mathbf{0.52} \text{ s} This is the time to reach the highest point.
The total time in the air is double this, since it takes the same time to go up and come back down: 2⋅0.52 s=1.04 s2 \cdot 0.52 \text{ s} = \mathbf{1.04} \text{ s}

STEP 9

Now that we know the ball is in the air for 1.04\mathbf{1.04} seconds, and its horizontal velocity is a constant 14.1 m/s\mathbf{14.1} \text{ m/s}, we can find the horizontal distance it travels!

STEP 10

Distance is just speed multiplied by time: Distance=Speed⋅Time\text{Distance} = \text{Speed} \cdot \text{Time} Distance=14.1 m/s⋅1.04 s≈14.7 m\text{Distance} = 14.1 \text{ m/s} \cdot 1.04 \text{ s} \approx \mathbf{14.7} \text{ m}

STEP 11

The ball will land approximately 14.7\mathbf{14.7} meters away from the person.

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