A long jumper leaves the ground at an angle of 25.2 to the horizontal and at a speed of 11.2 m/s.
How far does he jump? The acceleration
due to gravity is 9.8 m/s2 .
Answer in units of m
What maximum height does he reach?
A long jumper leaves the ground at an angle
of 23 ◦
to the horizontal and at a speed of
9.89 m/s.
How far does he jump? The acceleration
due to gravity is 9.8 m/s
2
.
Answer in units of m.
To find the horizontal distance the long jumper jumps, we can use the horizontal component of his velocity.
Step 1: Find the horizontal component of the velocity.
The horizontal component of the velocity can be found using the equation:
Vx = V * cos(theta)
where Vx is the horizontal component of velocity, V is the total velocity (11.2 m/s), and theta is the angle of takeoff (25.2 degrees).
Vx = 11.2 m/s * cos(25.2 degrees)
Vx ≈ 10.12 m/s (rounded to two decimal places)
Step 2: Find the time of flight.
The time of flight can be found using the equation:
t = (2 * Vy) / g
where t is the time of flight, Vy is the vertical component of velocity, and g is the acceleration due to gravity (9.8 m/s^2).
Since the long jumper leaves the ground with an angle of 25.2 degrees, the vertical component of velocity can be found using the equation:
Vy = V * sin(theta)
where Vy is the vertical component of velocity, V is the total velocity (11.2 m/s), and theta is the angle of takeoff (25.2 degrees).
Vy = 11.2 m/s * sin(25.2 degrees)
Vy ≈ 4.77 m/s (rounded to two decimal places)
t = (2 * 4.77 m/s) / 9.8 m/s^2
t ≈ 0.97 s (rounded to two decimal places)
Step 3: Calculate the horizontal distance.
The horizontal distance can be found using the equation:
d = Vx * t
where d is the horizontal distance, Vx is the horizontal component of velocity (10.12 m/s), and t is the time of flight (0.97 s).
d = 10.12 m/s * 0.97 s
d ≈ 9.82 m (rounded to two decimal places)
Therefore, the long jumper jumps approximately 9.82 meters.
To find the maximum height the long jumper reaches, we can use the vertical component of his velocity.
Step 4: Find the maximum height.
The maximum height can be found using the equation:
h = (Vy^2) / (2 * g)
where h is the maximum height, Vy is the vertical component of velocity (4.77 m/s), and g is the acceleration due to gravity (9.8 m/s^2).
h = (4.77 m/s)^2 / (2 * 9.8 m/s^2)
h ≈ 1.16 m (rounded to two decimal places)
Therefore, the long jumper reaches a maximum height of approximately 1.16 meters.
To find the distance the long jumper jumps, we can use the horizontal component of his velocity and the time he spends in the air.
The horizontal component of the velocity (Vx) can be found using trigonometry:
Vx = v * cosθ
Where v is the initial velocity (11.2 m/s in this case) and θ is the angle with respect to the horizontal (25.2° in this case).
Vx = 11.2 m/s * cos(25.2°)
Vx ≈ 10.057 m/s
The time the jumper spends in the air can be calculated using the vertical component of the velocity and the acceleration due to gravity:
Vy = v * sinθ
Where v is the initial velocity (11.2 m/s) and θ is the angle with respect to the horizontal (25.2°).
The time of flight (t) can be found using the formula:
t = 2 * Vy / g
Where g is the acceleration due to gravity (9.8 m/s²).
t = 2 * 11.2 m/s * sin(25.2°) / 9.8 m/s²
t ≈ 1.622 s
Now, we can calculate the distance jumped (D) using the horizontal component of the velocity and the time of flight:
D = Vx * t
D = 10.057 m/s * 1.622 s
D ≈ 16.316 m
So, the long jumper jumps approximately 16.316 meters.
To find the maximum height reached by the jumper, we can use the formula for projectile motion.
The maximum height (H) can be found using the vertical component of the velocity (Vy) and the time of flight (t):
H = Vy² / (2 * g)
H = (11.2 m/s * sin(25.2°))² / (2 * 9.8 m/s²)
H ≈ 1.100 m
So, the long jumper reaches a maximum height of approximately 1.100 meters.