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irakobra [83]
4 years ago
6

If the 140 g ball is moving horizontally at 25 m/s , and the catch is made when the ballplayer is at the highest point of his le

ap, what is his speed immediately after stopping the ball
Physics
1 answer:
sweet [91]4 years ago
7 0

(.140*25)+(75*0)= 3.5

3.5 = (.140 + 75)v

(3.5)/(.140 + 75) = v = .04657

which rounds up to .05 m/s

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kirill115 [55]

Answer:

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5 0
3 years ago
A river flows due east at 1.50 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
monitta

Answer:

a

   u =  10.11 \  m/s

b

   x = 44.996 \  m    

Explanation:

From the question we are told that    

        The velocity of the river due east is  v =  1.150i  \ m /s

        The velocity of the boat due north is  v_1 = 10j  \ m/ s

         The width of the river is  d =  300 \  m

Generally the velocity of the boat relative to the shore

           u =  \sqrt{v^2 + v_1^2}

=>        u =  \sqrt{1.50 ^2 + 10^2}

=>        u =  10.11 \  m/s

Generally direction of the boat relative to the shore is mathematically represented as

           tan \theta  =  \frac{v}{v_1}

=>        \theta  = tan^{-1}[ \frac{v}{v_1}]

=>        \theta  = tan^{-1}[ \frac{1.5}{10}]

=>        \theta  = 8.53^o

Generally the distance covered by the boat in the direction of flow of the river before reaching the north shore is mathematically represented as

             x =  d*  tan(\theta )

=>          x =  300 *  tan(8.53)

=>         x = 44.996 \  m          

         

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