Answer :
Answer:
Certainly, I can help you with this projectile motion problem.
To find the time the ball remains in the air, we can analyze its motion in the horizontal and vertical directions separately.
In the horizontal direction:
* The ball's initial horizontal velocity (v_x) is equal to its initial speed (v0) multiplied by the cosine of the launch angle (theta).
* v_x = v0 * cos(theta) = 20 m/s * cos(60°) = 10 m/s
Since there's no air resistance, the horizontal velocity remains constant throughout the flight.
In the vertical direction:
* The ball experiences constant downward acceleration due to gravity (g = 9.81 m/s²).
* We can use the initial height (h) and the vertical component of the initial velocity (v_y) to find the time (t) it takes for the ball to reach its maximum height.
* v_y = v0 * sin(theta) = 20 m/s * sin(60°) = 17.32 m/s (approx.)
Using the following kinematic equation:
* h = v_y * t - 0.5 * g * t²
where h = 2 m, v_y = 17.32 m/s (approx.), and g = 9.81 m/s²
Solving for t (the time to reach the maximum height):
* t = 2 * v_y / g = 2 * 17.32 m/s / 9.81 m/s² ≈ 3.53 s
Since the time it takes to fall back down to the ground is the same as the time it takes to reach the maximum height, the total time the ball remains in the air is approximately 2 * 3.53 s ≈ 7.06 s.
Therefore, the closest answer choice to the time the ball remains in the air is 3.53 seconds.
Answer:
3.64 s
Explanation:
The ball is a projectile in free fall. Because it undergoes constant acceleration, we can use kinematic equations also known as SUVAT equations to model its motion. In this case, we will use the equation:
[tex]\Large \text {$ s=ut+\frac{1}{2}at^2 $}[/tex]
where
- s is the displacement
- u is the initial velocity
- a is the acceleration
- t is the time
In the vertical direction, the ball's final position is 2 meters below its initial position, so its displacement is s = -2 m. The initial velocity is u = 20 m/s sin 60°. The acceleration is that due to gravity, a = -9.8 m/s². Plugging in these values:
[tex]\Large \text {$ s=ut+\frac{1}{2}at^2 $}\\\\\Large \text {$ -2=(20\sin60\textdegree)t+\frac{1}{2}(-9.8)t^2 $}\\\\\Large \text {$ -2=17.32t-4.9t^2 $}\\\\\Large \text {$ 4.9t^2-17.32t-2=0 $}[/tex]
Use the quadratic formula to solve for time.
[tex]\Large \text {$ t= $}\huge \text {$ \frac{-b\pm \sqrt{b^2-4ac} }{2a} $}\\\\\Large \text {$ t= $}\huge \text {$ \frac{17.32\pm \sqrt{(-17.32)^2-4(4.9)(-2)} }{2(4.9)} $}\\\\\Large \text {$ t= $}\huge \text {$ \frac{17.32\pm \sqrt{339.2} }{9.8} $}\\\\\Large \text {$ t=-0.112\ or\ t=3.65 $}[/tex]
Since the time can't be negative, t = 3.65, which is closest to 3.64 seconds.
