Vo = 89 m/s
angle: 40°
=> Vox = Vo * cos 40° = 89 * cos 40°
=> Voy = Vo. sin 40° = 89 * sin 40°
x-movement: uniform => x =Vox * t = 89*cos(40)*t
x = 300 m => t = 300m / [89m/s*cos(40) = 4.4 s
y-movement: uniformly accelerated => y = Voy * t - g*t^2 /2
y = 89m/s * sin(40) * (4.4s) - 9.m/s^2 * (4.4)^2 / 2 = 156.9 m = height the ball hits the wall.
Explanation:
(a)
The initial vertical velocity is 13 m/s. At the maximum height, the vertical velocity is 0 m/s.
v = at + v₀
0 = (-9.8) t + 13
t ≈ 1.33 s
(b)
Immediately prior to the explosion, the ball is at the maximum height. Here, the vertical velocity is 0 m/s, and the horizontal velocity is constant at 25 m/s.
v = √(vx² + vy²)
v = √(25² + 0²)
v = 25 m/s
(c)
Momentum is conserved before and after the explosion.
In the x direction:
m vx = ma vax + mb vbx
m (25) = (⅓ m) (0) + (⅔ m) (vbx)
25m = (⅔ m) (vbx)
25 = ⅔ vbx
vbx = 37.5 m/s
And in the y direction:
m vy = ma vay + mb vby
m (0) = (⅓ m) (0) + (⅔ m) (vby)
0 = (⅔ m) (vby)
vby = 0 m/s
Since the vertical velocity hasn't changed, and since Fragment B lands at the same height it was launched from, it will have a vertical velocity equal in magnitude and opposite in direction as its initial velocity.
vy = -13 m/s
And the horizontal velocity will stay constant.
vx = 37.5 m/s
The velocity vector is (37.5 i - 13 j) m/s. The magnitude is:
v = √(vx² + vy²)
v = √(37.5² + (-13)²)
v ≈ 39.7 m/s
They were both odd and obsessive
Explanation:
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