A ball is thrown toward a cliff of height h with a speed of 31 m/s and an angle of 60∘ above horizontal. It lands on the edge of the cliff 3.4 s later.
What is the ball's impact speed?
the horizontal velocity is
... 31 m/s * cos(60º)
the vertical velocity is
... 31 m/s * sin(60º) +
... 3.4 s * -9.8 m/s^2
add the two component vectors
... a^2 + b^2 = c^2
To find the ball's impact speed, we need to break down the problem into horizontal and vertical components.
1. Horizontal Component:
The horizontal component of the ball's initial velocity is given by:
Vx = V * cos(θ)
Where:
Vx = horizontal component of velocity
V = speed of the ball = 31 m/s
θ = angle above horizontal = 60 degrees
Substituting the given values:
Vx = 31 * cos(60)
Vx = 31 * 0.5
Vx = 15.5 m/s
2. Vertical Component:
The vertical component of the ball's initial velocity is given by:
Vy = V * sin(θ)
Where:
Vy = vertical component of velocity
V = speed of the ball = 31 m/s
θ = angle above horizontal = 60 degrees
Substituting the given values:
Vy = 31 * sin(60)
Vy = 31 * 0.866
Vy = 26.846 m/s
3. Time of Flight:
The time of flight for the ball is given as 3.4 seconds.
4. Vertical Displacement:
The vertical displacement of the ball is given by:
Δy = Vy * t + (1/2) * g * t^2
Where:
Δy = vertical displacement
Vy = vertical component of velocity = 26.846 m/s
t = time of flight = 3.4 s
g = acceleration due to gravity = 9.8 m/s^2 (assuming no air resistance)
Substituting the given values:
h = Vy * t + (1/2) * g * t^2
h = 26.846 * 3.4 + (1/2) * 9.8 * (3.4)^2
h = 91.3456 + 52.9544
h = 144.3 m
5. Impact Speed:
The impact speed of the ball is given by:
V_impact = sqrt(Vx^2 + Vy^2)
Where:
V_impact = impact speed
Vx = horizontal component of velocity = 15.5 m/s
Vy = vertical component of velocity = 26.846 m/s
Substituting the given values:
V_impact = sqrt((15.5)^2 + (26.846)^2)
V_impact = sqrt(240.25 + 720.6121)
V_impact = sqrt(960.8621)
V_impact = 31.00825 m/s (rounded to four decimal places)
Therefore, the ball's impact speed is approximately 31.0083 m/s.
To find the ball's impact speed, we first need to determine the vertical and horizontal components of its velocity at impact.
Given that the ball was thrown at an angle of 60 degrees above the horizontal, we can calculate the initial vertical velocity (Vy) and horizontal velocity (Vx) using trigonometric functions.
The initial vertical velocity can be found using the equation:
Vy = V * sin(θ)
where V is the initial speed and θ is the angle of elevation.
Vy = 31 m/s * sin(60°)
Vy = 31 m/s * 0.866
Vy ≈ 26.846 m/s
The horizontal velocity remains constant throughout the projectile motion, so the horizontal component of the velocity can be calculated using the equation:
Vx = V * cos(θ)
Vx = 31 m/s * cos(60°)
Vx = 31 m/s * 0.5
Vx = 15.5 m/s
Now let's focus on the vertical motion of the ball. We can use the height of the cliff and the time it takes for the ball to reach the edge of the cliff to find the vertical distance covered during this time.
We'll use the following kinematic equation to find the vertical distance covered:
h = Vy * t + (1/2) * g * t^2
where h is the height of the cliff, Vy is the vertical velocity, t is the time, and g is the acceleration due to gravity.
Since the ball lands on the edge of the cliff, the vertical displacement (h) is zero. Therefore, the equation becomes:
0 = Vy * t + (1/2) * g * t^2
Rearranging the equation, we can solve for t:
(1/2) * g * t^2 = - Vy * t
(1/2) * g * t = -Vy
t = (-2 * Vy) / g
Substituting the known values, we have:
t = (-2 * 26.846 m/s) / 9.8 m/s^2
t ≈ -5.46 s
Since time cannot be negative, the negative sign indicates that the ball landed 5.46 seconds before it was thrown. This suggests an error in the problem statement or calculation.
However, to find the ball's impact speed, we need the correct value of time. Once that is obtained, the impact speed can be calculated using the horizontal and vertical velocities at impact.
The magnitude of the impact velocity can be calculated using the Pythagorean theorem:
Impact speed = sqrt((Vx^2) + (Vy^2))
Substituting the known values:
Impact speed = sqrt((15.5 m/s)^2 + (26.846 m/s)^2)
Impact speed ≈ 31.04 m/s
Therefore, the ball's impact speed is approximately 31.04 m/s.