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melisa1 [442]
3 years ago
14

A baseball player (hitter) hits the ball with her bat. The ball flies north at an angle a above ground and with an initial speed

v towards a fielder whose job is to catch the ball before it hits the ground. The distance between the fielder and the hitter is D. At the moment when the bat hits the ball, the fielder starts running to the south towards the place where he believes the ball will fall. Assume the fielder runs with a constant speed all the time the ball is in the air. Assume the hand of the fielder catches the ball just before it hits the ground and that the ball is hit by the bat at ground level.
How fast must the fielder run to be able to catch the ball?
Physics
1 answer:
Vinvika [58]3 years ago
4 0

Answer:

imma try and still get wrong

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Answer:

a)  a=4\,\frac{m}{s^2}

b)  V(t)=4\,t\,+3

c)  V(1)=7 \,\frac{m}{s} \\

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e)  Average speed = 11 m/s

Explanation:

a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}

Therefore,  acceleration is a=4\,\frac{m}{s^2}

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

y=m\,x+b\\V(t)=4\,t\,+3

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = \frac{(7+15)\,2}{2} = 22\,\,m

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = \frac{22}{2} = 11\, \,\frac{m}{s}

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