A long jumper leaves the ground at an angle of 24.2 degrees to the horizontal and at a speed of 11 m/s.
A) How far does he jump?
B)What maximum height does he reach?
To find the answers to these questions, we can use the equations of motion for projectile motion. In projectile motion, the horizontal and vertical motion are independent of each other. We can break down the motion into horizontal and vertical components and solve for the desired quantities step by step.
A) How far does he jump?
To find the horizontal distance traveled by the long jumper, we need to determine the time of flight. The time of flight can be calculated using the initial vertical velocity and the acceleration due to gravity.
The initial vertical velocity (Vy0) can be found using the initial velocity (v0) and the angle of projection (θ) using the formula:
Vy0 = v0 * sin(θ)
Given that the initial velocity (v0) is 11 m/s and the angle of projection (θ) is 24.2 degrees, we can calculate the initial vertical velocity (Vy0):
Vy0 = 11 m/s * sin(24.2°) ≈ 4.45 m/s
We know that the vertical component of motion is influenced by gravity and follows the equation:
Vy = Vy0 + g * t
Where Vy is the vertical velocity, Vy0 is the initial vertical velocity, g is the acceleration due to gravity (approximated as 9.8 m/s²), and t is the time of flight.
At the highest point of the jump, the vertical velocity becomes zero. Thus, we can solve for the time of flight (t) using the equation:
0 = Vy0 + g * t
Rearranging this equation and solving for t:
t = -Vy0 / g
t = -4.45 m/s / (-9.8 m/s²) ≈ 0.454 seconds
Since the horizontal motion is uniform and unaffected by gravity, we can calculate the horizontal distance (d) using the formula:
d = v0 * cos(θ) * t
Plugging in the values, we find:
d = 11 m/s * cos(24.2°) * 0.454 s ≈ 4.88 meters
Therefore, the long jumper jumps approximately 4.88 meters horizontally.
B) What maximum height does he reach?
To find the maximum height reached by the long jumper, we can use the equation for vertical displacement:
Δy = Vy0 * t + (1/2) * g * t²
Where Δy is the vertical displacement, Vy0 is the initial vertical velocity, g is the acceleration due to gravity, and t is the time of flight.
We already found that the time of flight (t) is approximately 0.454 seconds, and the initial vertical velocity (Vy0) is 4.45 m/s.
Plugging in these values, we have:
Δy = 4.45 m/s * 0.454 s + (1/2) * 9.8 m/s² * (0.454 s)² ≈ 0.944 meters
Therefore, the long jumper reaches a maximum height of approximately 0.944 meters.