A long jumper leaves the ground at an angle of 24.4◦to the horizontal an data speed of 9.52m/s.How far does he jump?The acceleration due to gravity is 9.8m. What maximum height does he reach?
vertical problem first
Vi = 9.52 sin 24.4
v = Vi - 9.8 t
at top v = 0
so
9.8 t = 9.52 sin 24.4 = 3.93
t = .401 seconds to top
height
h = Vi t - 4.9 t^2
h = 3.93 (.401) - 4.9(.401)^2
h = .789 m
now the horizontal problem
time = 2 t = .802 seconds because he spends as much time falling as rising (same g)
u = 9.52 cos 24.4 forever
u = 8.67
x = u*time = 8.67 * .802 = 6.95 meters
To find the distance the long jumper jumps, we can use the horizontal component of the initial velocity. The horizontal velocity remains constant throughout the jump because there is no horizontal acceleration.
The horizontal component of the initial velocity (Vx) can be found using trigonometry:
Vx = initial velocity * cos(angle)
Vx = 9.52 m/s * cos(24.4°)
Vx = 9.52 m/s * 0.913
Vx = 8.69 m/s
Now, to find the distance, we can use the time of flight. The time of flight can be calculated by considering the vertical motion of the long jumper.
The vertical component of the initial velocity (Vy) can be found using trigonometry:
Vy = initial velocity * sin(angle)
Vy = 9.52 m/s * sin(24.4°)
Vy = 9.52 m/s * 0.394
Vy = 3.75 m/s
The time of flight (t) can be calculated using the vertical displacement (distance covered vertically):
Vertical displacement = (Vy * t) + (0.5 * acceleration due to gravity * t^2)
Since the jumper leaves and lands at the same vertical position, the vertical displacement is zero:
0 = (3.75 m/s * t) + (0.5 * 9.8 m/s^2 * t^2)
Simplifying the equation, we get a quadratic equation:
4.9 * t^2 + 3.75 * t = 0
Solving this equation, we get t = 0 or t = -0.765
We ignore the negative solution (t = -0.765) because time cannot be negative. Therefore, the time of flight (t) is 0 seconds.
Now, we can find the distance (d) using the horizontal velocity:
d = Vx * t
d = 8.69 m/s * 0 s
d = 0 m
So, the long jumper does not travel horizontally and therefore jumps a distance of 0 meters.
To find the maximum height reached by the long jumper, we can use the vertical component of the initial velocity (Vy) and the vertical displacement (distance covered vertically).
Using the same equation as before:
Vertical displacement = (Vy * t) + (0.5 * acceleration due to gravity * t^2)
With t = 0 s, the equation simplifies to:
Vertical displacement = 0 m
Therefore, the maximum height reached by the long jumper is 0 meters.