In the javelin throw at a track-and-field event, the javelin is launched at a speed of 33.0 m/s at an angle of 36.5° above the horizontal. How much time is required for it to hit the target that is the same height across the field?
the flight time is ...
2 [(launch velocity)(sine launch angle) / g]
To find the time required for the javelin to hit the target, we need to first determine the horizontal and vertical components of its velocity.
The horizontal component of velocity remains constant throughout the motion and can be calculated using the formula:
Vx = V * cos(θ)
where Vx is the horizontal component of velocity, V is the initial velocity of the javelin (33.0 m/s), and θ is the launch angle (36.5°).
Substituting the given values into the formula:
Vx = 33.0 m/s * cos(36.5°)
Vx ≈ 26.706 m/s
The vertical component of velocity changes as the javelin moves. The equation for the vertical motion of the javelin can be given as:
y = y0 + V0y * t - (1/2) * g * t^2
where y is the vertical displacement, y0 is the initial vertical position (same height as the target), V0y is the vertical component of velocity (upwards initially), t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the question specifies that the target is the same height across the field, the vertical displacement (y) is equal to zero.
0 = 0 + (33.0 m/s * sin(36.5°)) * t - (1/2) * (9.8 m/s^2) * t^2
Simplifying the equation:
16.5 * t - 4.9 * t^2 = 0
Now, we have a quadratic equation, and to solve for t, we need to set it equal to zero:
-4.9 * t^2 + 16.5 * t = 0
Factoring out a t:
t * (-4.9 * t + 16.5) = 0
Setting each term equal to zero:
t = 0 or -4.9 * t + 16.5 = 0
The first solution, t = 0, is not meaningful in the context of the problem since it represents the initial time when the javelin is released.
To find the non-zero solution:
-4.9 * t + 16.5 = 0
-4.9 * t = -16.5
t = -16.5 / -4.9
t ≈ 3.367 seconds
Therefore, it takes approximately 3.367 seconds for the javelin to hit the target when released at a speed of 33.0 m/s at an angle of 36.5° above the horizontal.