In the javelin throw at a track-and-field event, the javelin is launched at a speed of 35.0 m/s at an angle of 33.7 ° above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from 33.7 ° at launch to 16.3 °?

Vi = 35 sin 33.7 = 19.42 initial up

u = 35 cos 33.7 = 29.12 constant horizontal

v = Vi - g t
v = 19.42 - 9.8 t

when v/u = tan 16.3 we are there
v/29.12 = .2924
v = 8.515

so
8.515 = 19.42 - 9.8 t
t = 1.11 seconds

Compute the time that it takes for the vertical velocity component Vy to decrease from 35sin33.7 to 35cos33.7*tan16.3 .

(The horizontal velocity component remains 35cos33.7)

That time will be the difference of those numbers, divided by g.

(19.420 - 8.173)/9.81 = ___ s

To find the time required for the angle to decrease from 33.7° to 16.3°, we can use the equations of projectile motion.

The initial vertical velocity at launch can be calculated using the initial velocity (35.0 m/s) and the launch angle (33.7°). The vertical component of velocity (Vy) is given by:

Vy = V * sin(θ)

Where:
Vy = vertical component of velocity
V = initial velocity (35.0 m/s)
θ = launch angle (33.7°)

Substituting the values:

Vy = 35.0 * sin(33.7°)

Next, let's calculate the time (t1) required for the angle to decrease to 16.3°. At the point where the angle has decreased to 16.3°, the vertical component of velocity would be:

Vy_final = V * sin(16.3°)

Now, we can use the equation of motion for vertical displacement to find the time (t1):

Vy_final = Vy_initial - g*t1

Where:
Vy_final = final vertical component of velocity (V * sin(16.3°))
Vy_initial = initial vertical component of velocity (35.0 * sin(33.7°))
g = acceleration due to gravity (9.8 m/s², assuming no air resistance)
t1 = time required for the angle to decrease to 16.3°

Solving for t1:

V * sin(16.3°) = 35.0 * sin(33.7°) - 9.8 * t1

Now, we can rearrange the equation to solve for t1:

t1 = (35.0 * sin(33.7°) - V * sin(16.3°)) / 9.8

Substituting the known values:

t1 = (35.0 * sin(33.7°) - 35.0 * sin(16.3°)) / 9.8

Calculating the expression:

t1 ≈ 2.04 seconds

Therefore, it takes approximately 2.04 seconds for the angle to decrease from 33.7° to 16.3°.

To solve this problem, we need to analyze the motion of the javelin and find the time it takes for the angle to change.

Let's break down the problem into horizontal and vertical components.

Horizontal Component:
The initial velocity of the javelin in the horizontal direction remains constant. Therefore, the horizontal component of velocity does not change throughout the motion. We can find the horizontal component of velocity using the initial velocity and angle of launch.

Vx = V * cos(θ)
Vx = 35.0 m/s * cos(33.7°)

Vertical Component:
The vertical component of velocity changes over time due to the acceleration of gravity. The initial velocity in the vertical direction can be found using the initial velocity and angle of launch.

Vy = V * sin(θ)
Vy = 35.0 m/s * sin(33.7°)

The time to reach the highest point of the trajectory is given by the formula:

t = Vy / g

where g is the acceleration due to gravity, approximately 9.8 m/s².

After reaching the highest point, the angle decreases symmetrically until it reaches 16.3°.

To find the time it takes for the angle to decrease from 33.7° to 16.3°, we need to consider only the vertical component of velocity.

Using the formula for vertical velocity:

Vy = Vy_initial - g * t

We know the initial vertical velocity (Vy_initial) and can calculate it using the given information:

Vy_initial = V * sin(θ)
Vy_initial = 35.0 m/s * sin(33.7°)

Now we can set up the equation to find the time for the angle to decrease:

16.3° = arcsin((Vy_initial - g * t) / V)

Solving this equation will give us the value of t, which represents the time it takes for the angle to decrease from 33.7° to 16.3°.