A long jumper leaves the ground at an angle
of 24.3◦ to the horizontal and at a speed of 10.6 m/s.
How far does he jump?
To find the horizontal distance the long jumper jumps, we need to consider the horizontal component of the initial velocity.
Step 1: Calculate the horizontal component of the initial velocity using the formula:
Vx = V * cos(θ)
where Vx is the horizontal component of the initial velocity, V is the initial velocity magnitude (10.6 m/s), and θ is the angle (24.3 degrees).
Vx = 10.6 m/s * cos(24.3°)
Step 2: Calculate the time the jumper is in the air.
Since there is no vertical acceleration during the jump, we can use the following equation:
Δy = Vy * t + (1/2) * g * t^2
Where Δy is the maximum height, Vy is the vertical component of the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.
At the maximum height, Vy = 0 m/s, and Δy can be assumed to be negligible compared to the horizontal distance. Thus, we can simplify the equation to:
0 = (1/2) * g * t^2
Solving for t:
(1/2) * g * t^2 = 0
t = 0 or t = √0
Since t cannot be zero, we have t = 0 s.
Step 3: Calculate the horizontal distance using the formula:
Δx = Vx * t
Substituting the values:
Δx = (10.6 m/s * cos(24.3°)) * 0 s
Since t is zero, the horizontal distance is also zero.
Therefore, the long jumper does not jump any distance horizontally and remains at the same spot.
To find the distance the long jumper jumps, we need to analyze the horizontal and vertical components of their motion.
1. Horizontal Component:
The horizontal motion of the jumper is not affected by the angle of takeoff. It can be treated as a projectile motion with constant velocity.
The horizontal component of the jumper's velocity is given by:
Vx = V * cos(θ)
Vx = 10.6 m/s * cos(24.3◦)
Vx ≈ 10.6 m/s * 0.907
Vx ≈ 9.62 m/s (rounded to two decimal places)
2. Vertical Component:
The vertical motion of the jumper is influenced by the takeoff angle and is governed by the laws of projectile motion.
The vertical component of the jumper's velocity is given by:
Vy = V * sin(θ)
Vy = 10.6 m/s * sin(24.3◦)
Vy ≈ 10.6 m/s * 0.409
Vy ≈ 4.34 m/s (rounded to two decimal places)
Using the vertical component, we can find the time of flight:
t = (2 * Vy) / g
where g is the acceleration due to gravity and is approximately 9.8 m/s^2.
t = (2 * 4.34 m/s) / 9.8 m/s^2
t ≈ 0.885 seconds (rounded to three decimal places)
Now, we can calculate the horizontal distance:
d = Vx * t
d = 9.62 m/s * 0.885 s
d ≈ 8.51 meters (rounded to two decimal places)
Therefore, the long jumper jumps approximately 8.51 meters.