A longjumper running at 12 m/s and jumps at 20 degrees. No wind or resistance but gravity is normal -9.81 m/s2. How far will person jump?
Figure his time in air first.
Because time in air is determined by gravity, consider the vertical first.
Vertical: hf=ho+Vi*t-4.9t^2
0=0+12Sin20*t-4.9t^2
solve for time in air, t.
Now, the problem, how far does he go.
distance=horizontalvelocity*time
= 12cos20*timeinair
Ok..I am not sure how to solve for t?
You may need to drop physics, the language of physics is math, and this is a pretty basic algebra I skill.
0=0+12sin20*t-4.9t^2
0=t(12sin20-4.9t)
so either t is zero, or t is 12sin20/4.9
To calculate the distance the long jumper will jump, we can break down the motion of the jumper into horizontal and vertical components.
First, let's find the vertical component of the jumper's velocity. We can use the equation:
v = u + at
where:
v = final velocity (vertical component) after the jump (0 m/s, as the jumper comes to a stop at the peak of the jump)
u = initial velocity (vertical component) before the jump (0 m/s, as the jumper is not initially moving vertically)
a = acceleration due to gravity (-9.81 m/s^2)
t = time taken to reach the peak of the jump (which we can find using the formula v = u + at)
Since the jumper comes to a stop at the peak of the jump (final vertical velocity is 0 m/s), we can use this equation to solve for the time taken to reach the peak:
0 = 0 + (-9.81) * t
Rearranging the equation, we get:
t = 0 / (-9.81) = 0 seconds
Therefore, the time taken to reach the peak is 0 seconds.
Next, let's find the horizontal component of the jumper's velocity. The horizontal component of the jumper's velocity remains constant throughout the jump, as there is no horizontal force acting on the jumper. Therefore, the horizontal component of the jumper's initial velocity (u) also remains constant.
To find the horizontal component of the velocity (ux), we can use the equation:
ux = u * cosθ
where:
u = initial velocity magnitude (12 m/s)
θ = angle at which the jumper jumps (20 degrees)
Substituting the given values into the equation, we get:
ux = 12 m/s * cos(20 degrees) ≈ 11.10 m/s
Now, we can calculate the horizontal distance traveled (S) using the equation:
S = ux * t
where:
S = horizontal distance traveled by the jumper
ux = horizontal component of the velocity (11.10 m/s)
t = time taken to reach the peak of the jump (0 seconds)
Substituting the values into the equation, we get:
S = 11.10 m/s * 0 seconds = 0 meters
Therefore, considering the given conditions, the person will not move horizontally and will jump a distance of 0 meters.