An Olympic long jumper is capable of jumping 8.44m. Assuming his horizontal speed is 5.644m/s as he leaves the ground, how high does he go?
time in air:
distance=vh*time
time= 8.44/5.644 sec
so max heigh give a falling time 1/2 time in air.
hf=hmax-1/2 *9.8*(1/2*8.44/5.644)^2
hf=0, solve for hmax
To determine the height that an Olympic long jumper reaches, we need to make use of the principles of projectile motion. In this case, we have the horizontal velocity and the horizontal distance traveled by the jumper. We need to find the vertical distance or height reached by the jumper.
Let's break down the problem into two components: horizontal motion and vertical motion.
1. Horizontal Motion:
Since we have the horizontal speed of 5.644 m/s, we know that it remains constant throughout the motion. Therefore, we can use the equation:
distance = speed × time
In this case, the distance is the horizontal distance traveled, which is given as 8.44 m. The speed is 5.644 m/s. Hence, we can rearrange the equation and solve for time:
time = distance / speed
time = 8.44 m / 5.644 m/s
time ≈ 1.494 seconds
2. Vertical Motion:
In vertical motion, we need to consider the height reached by the jumper. We can use the following kinematic equation to calculate the height:
height = initial vertical velocity × time + (1/2) × acceleration × time^2
In this scenario, we assume the initial vertical velocity (vertical speed) to be zero, as the jumper starts from the ground. The acceleration due to gravity (g) is approximately 9.8 m/s^2 (neglecting air resistance). Plugging in the values, we have:
height = 0 × 1.494 + (1/2) × 9.8 × (1.494)^2
height ≈ 11.07 meters
Therefore, the Olympic long jumper reaches a height of approximately 11.07 meters.