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?
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.
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.
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