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Mazyrski [523]
3 years ago
12

g An arrow is shot straight up in the air at an initial speed of 15.5 m/s. After how much time is the arrow moving downward at a

speed of 5.20 m/s
Physics
1 answer:
Sati [7]3 years ago
5 0

Answer:

2.11 seconds

Explanation:

We use the kinematic equation for the velocity in a constantly accelerated motion under the acceleration of gravity (g):

v_f=v_i-g*t\\-5.2 = 15.5 - 9.8\,t\\9.8 \,t= 15.5 + 5.2\\t = 20.7/9.8\\t = 2.11 \,\,sec

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Answer:

x = 0.54 m

y = 0.058 m

Explanation:

m = mass of the bullet = 16 g = 0.016 kg

v = speed of bullet before collision = 240 m/s

M = mass of the pendulum = 3.6 kg

L = length of the string = 2.5 m

h = height gained by the pendulum after collision

V = speed of the bullet and pendulum combination

Using conservation of momentum

m v = (m + M) V

(0.016) (240) = (0.016 + 3.6) V

V = 1.062 m/s

Using conservation of energy

Potential energy gained by bullet and pendulum combination = Kinetic energy of bullet and pendulum combination

(m + M) g h = (0.5) (m + M) V²

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y = vertical displacement = h = 0.058 m

x = horizontal displacement

horizontal displacement is given as

x = sqrt(L² - (L - h)²)

x =  sqrt(2.5² - (2.5 - 0.058)²)

x = 0.54 m

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