Intial velocity u=3m/s
final velocity v
2
=u
2
+2as=3
2
+(2×2×5)=29 ⟹v=5.3m/s
KE=
2
1
m(v
2
−u
2
)=
2
1
×2×((5.3)
2
−3
2
)=20J
It gets heavier and heavier
Answer:
F = 1176 N
Explanation:
Given that,
Mass of a baseball, m = 0.140-kg
Initial speed of the baseball, u = 33.6 m/s
Final speed of the baseball, v = -33.6 m/s (in opposite direction)
The time of contact of the bat and the ball, t = 0.008 s
We need to find the average force the bat exerts on the ball. The force acting on the ball is given by :

So, the average force the bat exerts on the ball is 1176 N.
Answer:
a) Vb = 4.65m/s at 25.46° due south of east
b) t = 119s
c) d = 238m
d) 28.43° due north of east.
e) Vb = 3.69m/s due east
f) t = 135.5s
Explanation:
Velocity relative to the earth is:
![V_b = V_{b/w} + V_w = [4.2,0]+[0,-2]=[4.2,-2]m/s](https://tex.z-dn.net/?f=V_b%20%3D%20V_%7Bb%2Fw%7D%20%2B%20V_w%20%3D%20%5B4.2%2C0%5D%2B%5B0%2C-2%5D%3D%5B4.2%2C-2%5Dm%2Fs)
Vb = 4.65m/s < -25.46°
Since the distance to travel is 500m:

The distance on the y-axis is given by:

If the final position is directly east from the starting position:

![V_b= [V_{b-x}, 0] = [4.2*cos\beta ,4.2*sin\beta ]+[0,-2]](https://tex.z-dn.net/?f=V_b%3D%20%5BV_%7Bb-x%7D%2C%200%5D%20%3D%20%5B4.2%2Acos%5Cbeta%20%2C4.2%2Asin%5Cbeta%20%5D%2B%5B0%2C-2%5D)
From the y-components:
Solving for β:
β=28.43°
With this angle, the velocity would be:
Vb = 4.2*cos(28.43°) = 3.69m/s
And the time it would take it to cross:
t = 500/3.69 = 135.5s
Answer:
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Explanation:
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