This is about tennis and doing a lob when an opponent is near the net. Suppose t
ID: 1952794 • Letter: T
Question
This is about tennis and doing a lob when an opponent is near the net.Suppose that you loft the ball with an initial speed of 15 m/s at an angle of 50 degrees above the horizontal. At this instant your opponent is 10 m away from the ball. he begins moving away from you 0.30 seconds later, hoping to reach the ball and hit it back at the moment that it is 2.10 m above its launch point. with what minimum average speed must he move? (Ignore the fact that he can stretch so that his racket can reach the ball before he does)
I made a diagram on the problem to get a clearer view on it: http://i125.photobucket.com/albums/p80/bright_claire/tennisquestion.jpg (copy paste url at a new tab). the green dot is just the tennis ball.
Explanation / Answer
I am going to guess that the launch point is so close to the ground that it is practically on the ground and that the 2.10 m above the lunch point is 2.10 m above the ground. Initial speed is 15.0 m/s at 50.0 degrees above the horizontal. The vertical component of the initial velocity (Vo); Vo=15sin(50 deg)=11.5m/s The horizontal component = 15cos(50 deg)=9.6 m/s The time the ball takes on this journey is the time it takes to reach its peak and fall to the ground until it is 2.10 m above the ground. Time it takes to peak is found from the following; velocity=initial velocity- accel of gravity*t At its peak V=0 0=11.5-9.81t solve for t and you get t=1.17 m/s The height is found from this equation; height=initial height +Vot-1/2 accel of gravity*t^2 height=0+11.5(1.17)-1/2(9.81)(1.17)^2=… meters The only force to bring it down is gravity. To make the ball fall the distance of 6.74 to 2.10 meters (4.64 meters) is found from the following; distance= 1/2accel of gravity t^2 4,64=1/2(9.81)t^2 Solve for t and you get t= 0.97 + 1.17 =2.14 sec. The Distance the ball goes is from the following eqn; distance=velocity*time distance=9.6(2.14)=20.6 meters This player has a reaction time of 0.36 sec before he starts to go back. The time is 2.14 sec-0.36 sec=1.78 sec The speed the guy must go is change in distance/ change in time speed=20.6 meters/1.78 sec=11.5 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.