A long jumper leaves the ground at an angle of 22.7◦ to the horizontal and at a speed of 9.47 m/s.
How far does he jump? The acceleration due to gravity is 9.8 m/s2 .
Answer in units of m
To find how far the long jumper jumps, we can use the equations of motion.
First, let's break down the initial velocity of the jumper into its horizontal and vertical components. The horizontal component can be found by multiplying the initial speed (9.47 m/s) by the cosine of the angle (22.7°):
Horizontal component = initial speed × cos(angle)
Horizontal component = 9.47 m/s × cos(22.7°)
Similarly, the vertical component can be found by multiplying the initial speed (9.47 m/s) by the sine of the angle (22.7°):
Vertical component = initial speed × sin(angle)
Vertical component = 9.47 m/s × sin(22.7°)
Now, let's calculate the time taken for the jumper to reach the maximum height. At the maximum height, the vertical component of the velocity becomes zero. Using the equation:
Vertical component = initial vertical velocity + (acceleration due to gravity × time)
0 = 9.47 m/s × sin(22.7°) + (9.8 m/s^2 × time)
0 = 2.15 m/s + (9.8 m/s^2 × time)
Solving for "time," we find:
9.8 m/s^2 × time = -2.15 m/s
time = -2.15 m/s / 9.8 m/s^2
Now that we have the time taken to reach maximum height, we can use this time to calculate the total time of flight. Since the jumper reaches the maximum height twice during the jump, the total time of flight is twice the time taken to reach the maximum height.
Total time of flight = 2 × time
Once we have the total time of flight, we can multiply it by the horizontal component of velocity to find the horizontal distance the jumper travels:
Distance = horizontal component × total time of flight
Now you can substitute the values into the equations to find the distance the long jumper jumps.