you throw a ball from a cliff with an initial velocity of 15m/s at an angle of 20° below the horizontal.find(a) its horizontal displacement and(b) its vertical displacement 2.3sec later?

Question

you throw a ball from a cliff with an initial velocity of 15m/s at an angle of 20° below the horizontal.find(a) its horizontal displacement and(b) its vertical displacement 2.3sec later?

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3 answers
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Answer 1

MathMate answered
9 years ago

Resolve the initial velocities into the

horizontal component, vx =v0 cos(θ)
and the
vertical component, vy= v0 sin(θ)
where θ=-20 degrees (below horiz.)

The horizontal component is subjected to only air resistance which in general is neglected, so travels at a uniform velocity.

The vertical component is subjected to the acceleration due to gravity, so use the formula
Δx=x0+v0(t)+(1/2)gt^2
x0=initial displacement = 0
v0=initial vertical velocity as given above.
g=acceleration due to gravity = -9.81 m/s^2

Answer 2

Step-by-Step Bot answered
1 year ago

To find the horizontal displacement and vertical displacement of the ball after 2.3 seconds, we can use the equations of projectile motion.

Given:
Initial velocity (v₀) = 15 m/s
Launch angle (θ) = 20° below the horizontal
Time (t) = 2.3 seconds

Step 1: Break down the initial velocity into horizontal and vertical components.

The horizontal component of velocity (v₀x) is given by:
v₀x = v₀ * cos(θ)
v₀x = 15 * cos(20°)
v₀x ≈ 14.20 m/s

The vertical component of velocity (v₀y) is given by:
v₀y = v₀ * sin(θ)
v₀y = 15 * sin(20°)
v₀y ≈ 5.11 m/s

Step 2: Calculate the horizontal displacement (x):

The horizontal displacement is given by:
x = v₀x * t
x = 14.20 * 2.3
x ≈ 32.66 meters

Therefore, the horizontal displacement of the ball after 2.3 seconds is approximately 32.66 meters.

Step 3: Calculate the vertical displacement (y):

The vertical displacement is given by the equation of motion:
y = v₀y * t + (1/2) * g * t²
Note: g represents the acceleration due to gravity, which is approximately 9.8 m/s².

y = 5.11 * 2.3 + (1/2) * 9.8 * (2.3)²
y ≈ 11.75 meters

Therefore, the vertical displacement of the ball after 2.3 seconds is approximately 11.75 meters.

To summarize:
(a) The horizontal displacement is approximately 32.66 meters.
(b) The vertical displacement is approximately 11.75 meters.

Answer 3

Explain Bot answered
1 year ago

To find the horizontal displacement and vertical displacement of a ball thrown from a cliff, we can use the equations of motion.

The horizontal displacement can be found using the formula:
Horizontal Displacement = Initial velocity * Time * Cos(angle)

The vertical displacement can be found using the formula:
Vertical Displacement = (Initial velocity * Time * Sin(angle)) + (0.5 * Acceleration * Time^2)

Given:
Initial velocity (Vi) = 15 m/s
Angle (θ) = 20° below the horizontal
Time (t) = 2.3 sec

First, let's convert the angle from degrees to radians, as trigonometric functions typically use radians.

θ = 20° * (π / 180°) ≈ 0.3491 radians

Now, we can calculate the horizontal displacement (a) and vertical displacement (b):

(a) Horizontal Displacement = Vi * t * Cos(θ)
Horizontal Displacement = 15 m/s * 2.3 sec * Cos(0.3491)
Horizontal Displacement ≈ 30.0694 m

(b) Vertical Displacement = (Vi * t * Sin(θ)) + (0.5 * g * t^2)
We need the acceleration due to gravity (g), which is approximately 9.8 m/s^2
Vertical Displacement = (15 m/s * 2.3 sec * Sin(0.3491)) + (0.5 * 9.8 m/s^2 * (2.3 sec)^2)
Vertical Displacement ≈ 6.65 m

Therefore, the horizontal displacement is approximately 30.0694 meters, and the vertical displacement 2.3 seconds later is approximately 6.65 meters.