A long jumper leaves the ground at an angle of 22.1◦ to the horizontal and at a speed of 10 m/s.
How far does he jump? The acceleration due to gravity is 9.8 m/s2 .
Answer in units of m
Well, let's calculate it step by step. First, we need to find the vertical and horizontal components of the initial velocity.
The vertical component can be found using the formula: Vy = V * sin(theta), where V is the initial velocity and theta is the launch angle.
Vy = 10 m/s * sin(22.1◦) ≈ 4.0 m/s
Now, we can find the time it takes for the jumper to reach the highest point of the jump. Since the vertical velocity changes due to gravity, we'll use the formula:
Vy = Vy0 + g * t
Where Vy0 is the initial vertical velocity, g is the acceleration due to gravity, and t is time. At the highest point, Vy will be zero.
0 = 4.0 m/s - 9.8 m/s² * t
Solving for t:
t = 4.0 m/s / (9.8 m/s²) ≈ 0.41 s
Now, let's find the time it takes for the jumper to fall back to the ground. We'll use the same formula, but this time with an initial vertical velocity of zero.
0 = 0 + 9.8 m/s² * t
Solving for t:
t = 0 s (that makes sense, since at the highest point Vy was zero)
Now, we can find the horizontal distance traveled using the formula:
distance = Vx * t
Since there is no horizontal acceleration, the initial horizontal velocity remains constant at the initial velocity.
distance = 10 m/s * 0.41 s ≈ 4.1 m
So, the long jumper jumps a distance of approximately 4.1 meters.
Now, that's why they say jumping to conclusions isn't always the best idea!
To find the distance the long jumper jumps, we can use the equations of projectile motion. The horizontal motion can be considered uniform as there is no force acting horizontally. The vertical motion can be considered as accelerated motion due to gravity.
First, let's break the initial velocity into horizontal and vertical components.
Given:
Angle of takeoff (θ) = 22.1°
Initial speed (v₀) = 10 m/s
Acceleration due to gravity (g) = 9.8 m/s²
To find the vertical component of the initial velocity (v₀y), we use:
v₀y = v₀ * sin(θ)
v₀y = 10 m/s * sin(22.1°)
v₀y ≈ 3.73 m/s
The time it takes for the jumper to reach the maximum height can be found using the vertical component of velocity (v₀y) and the acceleration due to gravity (g):
t = v₀y / g
t ≈ 3.73 m/s / 9.8 m/s²
t ≈ 0.38 s
Since the motion is symmetrical, the total time of flight (T) is twice the time it takes to reach maximum height:
T = 2 * t
T ≈ 2 * 0.38 s
T ≈ 0.76 s
Now, we can find the horizontal distance traveled using the horizontal component of velocity (v₀x) and the time of flight (T):
v₀x = v₀ * cos(θ)
v₀x = 10 m/s * cos(22.1°)
v₀x ≈ 9.29 m/s
d = v₀x * T
d ≈ 9.29 m/s * 0.76 s
d ≈ 7.06 m
Therefore, the long jumper jumps approximately 7.06 meters.
To find how far the long jumper jumps, we can use projectile motion equations. The horizontal distance traveled by the long jumper can be calculated using the formula:
x = v * t
Where:
x = horizontal distance traveled
v = horizontal component of velocity
t = time of flight
First, let's find the horizontal component of velocity. We can use trigonometry to calculate it:
v_horizontal = v * cos(theta)
Where:
v = initial velocity
theta = angle of takeoff
Substituting the given values:
v_horizontal = 10 m/s * cos(22.1°)
Now, let's find the time of flight. The time of flight can be determined using the formula:
t = 2 * v * sin(theta) / g
Where:
v = initial velocity
theta = angle of takeoff
g = acceleration due to gravity
Substituting the given values:
t = 2 * 10 m/s * sin(22.1°) / 9.8 m/s²
Now, we can substitute the calculated values into the horizontal distance formula:
x = (10 m/s * cos(22.1°)) * (2 * 10 m/s * sin(22.1°) / 9.8 m/s²)
Simplifying the equation will give us the final answer in units of meters.