Question

A long jumper leaves the ground at an angle of 20 degrees above the horizontal at a speed of 11 m/s. How far does he jump with a result all direction? What is the maximum height reached?

Answer 1

To find the distance the long jumper jumps and the maximum height reached, we can use the equations of projectile motion. Let's assume there is no air resistance.

Given: Launch angle, θ = 20 degrees Initial speed, v0 = 11 m/s Acceleration due to gravity, g ≈ 9.8 m/s² (approximate value on Earth)

1. Finding the horizontal distance: The horizontal distance traveled by the long jumper can be calculated using the formula:

Horizontal distance = (Initial speed * time of flight) * cos(θ)

The time of flight can be calculated using the formula:

Time of flight = (2 * Initial speed * sin(θ)) / g

Plugging in the given values: Time of flight = (2 * 11 m/s * sin(20°)) / 9.8 m/s²

Now, we can substitute the time of flight into the horizontal distance formula:

Horizontal distance = (11 m/s * [(2 * 11 m/s * sin(20°)) / 9.8 m/s²]) * cos(20°)

2. Finding the maximum height: The maximum height reached by the long jumper can be calculated using the formula:

Maximum height = (Initial speed * sin(θ))^2 / (2 * g)

Plugging in the given values: Maximum height = (11 m/s * sin(20°))^2 / (2 * 9.8 m/s²)

Solving these equations will give us the desired results.