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mac227
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projectile problem help!
The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net (see the drawing). Suppose that you loft the ball with an initial speed of 15.0 m/s at an angle of 50.0° the horizontal. At this instant your opponent is 10.0 m away from the ball. He begins moving away from you 0.27 s 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.)
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)=11.49m/s
The horizontal component = 15cos(50 deg)=9.64 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=11.49-9.81t
solve for t and you get t=1.1919m/s
The height is found from this equation;
height=initial height +Vot-1/2 accel of gravity*t^2
height=0+11.49(1.1919)-1/2(9.81)(1.1919)^2= 6.7267 meters
The only force to bring it down is gravity. To make the ball fall the distance of 6.7267 to 2.10 meters (4.6276 meters) is found from the following;
distance= 1/2accel of gravity t^2
4.6276=1/2(9.81)t^2
Solve for t and you get t= 0.9716 + 1.1919 =2.1635 sec.
The Distance the ball goes is from the following eqn;
distance=velocity*time
distance=9.64(2.1635)=20.85614 meters
This player has a reaction time of 0.27 sec before he starts to go back. The time is 2.1635 sec-0.27 sec=1.8935 sec
The speed the guy must go is change in distance/ change in time
speed=20.85614 meters/1.8935 sec=11.014m/s
THe online homework says I'm wrong.
Where did I go Wrong!?
Homework Statement
The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net (see the drawing). Suppose that you loft the ball with an initial speed of 15.0 m/s at an angle of 50.0° the horizontal. At this instant your opponent is 10.0 m away from the ball. He begins moving away from you 0.27 s 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.)
Homework Equations
The Attempt at a Solution
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)=11.49m/s
The horizontal component = 15cos(50 deg)=9.64 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=11.49-9.81t
solve for t and you get t=1.1919m/s
The height is found from this equation;
height=initial height +Vot-1/2 accel of gravity*t^2
height=0+11.49(1.1919)-1/2(9.81)(1.1919)^2= 6.7267 meters
The only force to bring it down is gravity. To make the ball fall the distance of 6.7267 to 2.10 meters (4.6276 meters) is found from the following;
distance= 1/2accel of gravity t^2
4.6276=1/2(9.81)t^2
Solve for t and you get t= 0.9716 + 1.1919 =2.1635 sec.
The Distance the ball goes is from the following eqn;
distance=velocity*time
distance=9.64(2.1635)=20.85614 meters
This player has a reaction time of 0.27 sec before he starts to go back. The time is 2.1635 sec-0.27 sec=1.8935 sec
The speed the guy must go is change in distance/ change in time
speed=20.85614 meters/1.8935 sec=11.014m/s
THe online homework says I'm wrong.
Where did I go Wrong!?