In the javelin throw at a track-and-field event, the javelin is launched at a speed of 33 m/s at an angle of 32° 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 32° at launch to 21°?
To find the time required for the angle to decrease from 32° to 21°, we need to first find the initial vertical velocity of the javelin at launch.
Given:
Initial speed, v₀ = 33 m/s
Launch angle, θ₀ = 32° above the horizontal
Step 1: Resolve the initial velocity components
The initial horizontal velocity, v₀x, remains constant throughout the javelin's flight. Therefore, we have:
v₀x = v₀ * cos(θ₀)
The initial vertical velocity, v₀y, can be calculated using:
v₀y = v₀ * sin(θ₀)
Step 2: Find the time it takes for the angle to decrease from 32° to 21°
Let's assume the time for the angle to decrease from 32° to 21° is t seconds.
To find the vertical velocity, v_y, as a function of time, we use the following equation of motion:
v_y = v₀y - g * t
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Step 3: Calculate the new launch angle
The new launch angle, θ, is given by the expression:
θ = arctan(v_y / v_x)
where v_x is the constant horizontal velocity.
Step 4: Solve for time t
We can substitute the expressions for θ, v_y, and v_x into the equation above to find t. Rearranging the equation, we have:
t = (v₀y - v₀ * tan(θ) ) / g
Substituting the known values, we can solve for t:
t = (v₀ * sin(θ₀) - v₀ * tan(21°) ) / g
Finally, we can calculate the value of t to find out the time required for the angle to decrease from 32° to 21°.
To determine how much time is required for the angle of the javelin to decrease from 32° at launch to 21°, we need to consider the projectile motion of the javelin.
Projectile motion is the motion of an object that is launched into the air and follows a curved path under the influence of gravity. In this case, the javelin is launched at an angle above the horizontal, so we will need to break down its initial velocity into horizontal and vertical components.
First, let's find the initial vertical velocity (Viy) using the given launch speed and launch angle. We can use trigonometric functions to find this component:
Viy = V * sin(θ)
= 33 m/s * sin(32°)
Now, we can find the time it takes for the angle to decrease from 32° to 21° by using the equation:
θf = atan(Viyf / Vx)
where θf is the final angle, Viyf is the final vertical velocity, and Vx is the initial horizontal velocity. Rearranging the equation, we have:
Viyf = Vx * tan(θf)
We know Viyf is equal to Viy because there is no horizontal acceleration. Substituting the values:
V * sin(21°) = V * sin(32°) * tan(21°)
Now we can solve for time (t) by replacing Viyf with Viy:
Viy = Viyf
V * sin(32°) = V * sin(21°) * tan(21°)
Taking the ratio of the two equations above, we can eliminate the velocity (V) term:
(sin(32°) / sin(21°)) = tan(21°)
We can solve this equation to find the ratio between the initial and final angles. Taking the inverse tangent of both sides:
θratio = arctan((sin(32°) / sin(21°)))
Now we can use this ratio to determine the time required for the angle to decrease. Since we have the initial angle (θi) and the ratio (θratio), we can calculate the time:
θi - θf = θratio * t
Rearranging the equation to solve for time:
t = (θi - θf) / θratio
Plugging in the values:
t = (32° - 21°) / θratio
Finally, we can substitute the value of θratio and calculate the time required for the angle to decrease from 32° to 21°.