a soccer player is practicing on the field. he kicks a ball with a speed of 32 m/s at a 53 - degree angle…

a soccer player is practicing on the field. he kicks a ball with a speed of 32 m/s at a 53 - degree angle with respect to the horizontal toward the net. if the horizontal bar of the goal is 2.3 meters above the ground and he is standing 35 - yards in from of the goal, does he score a goal or does the ball soar above the net? answer this question by finding the height of the ball above the ground when it reaches the net.
Answer
Explanation:
Step1: Convert yards to meters
1 yard = 0.9144 meters, so 35 yards = 35×0.9144 = 32.004 meters.
Step2: Find the initial horizontal and vertical velocities
The initial velocity $v_0 = 32$ m/s. The initial horizontal velocity $v_{0x}=v_0\cos\theta$, where $\theta = 53^{\circ}$, so $v_{0x}=32\cos53^{\circ}=32\times0.6018 = 19.2576$ m/s. The initial vertical velocity $v_{0y}=v_0\sin\theta=32\sin53^{\circ}=32\times0.7986 = 25.5552$ m/s.
Step3: Find the time it takes for the ball to reach the net
We know that in the horizontal - direction $x = v_{0x}t$. So $t=\frac{x}{v_{0x}}$, substituting $x = 32.004$ m and $v_{0x}=19.2576$ m/s, we get $t=\frac{32.004}{19.2576}\approx1.662$ s.
Step4: Find the height of the ball at time t
The equation for the vertical displacement is $y = v_{0y}t-\frac{1}{2}gt^{2}$, where $g = 9.8$ m/s². Substituting $v_{0y}=25.5552$ m/s, $t = 1.662$ s and $g = 9.8$ m/s², we have $y=25.5552\times1.662-\frac{1}{2}\times9.8\times(1.662)^{2}$. $y = 42.4727 - 13.4977=28.975$ m.
Answer:
The ball soars above the net since the height of the ball when it reaches the net is approximately 28.98 m which is much higher than the 2.3 - meter - high horizontal bar of the goal.