A 139 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 89 m below
1 answer:
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Answer:
5.4 ms⁻¹
Explanation:
Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.
= length of the meter stick = 1 m
= mass of the meter stick
= angular speed of the meter stick as it hits the floor
= speed of the other end of the stick
we know that, linear speed and angular speed are related as
= height of center of mass of meter stick above the floor =
= Moment of inertia of the stick about one end
For a stick, momentof inertia about one end has the formula as
Using conservation of energy
Rotational kinetic energy of the stick = gravitational potential energy
Answer:
a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m
Explanation:
a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.
So y - y₀ = ut - 1/2gt²
y - y₀ = (v₀sinθ)t - 1/2gt²
substituting the values of the variables into the equation, we have
0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²
- 60 = 50t - 4.9t²
So, 4.9t² - 50t - 60 = 0
Using the quadratic formula to find t,
Since t cannot be negative, t = 11.29 s
b. We first need to find the impact vertical velocity component. Using
v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,
v = v₀sinθ - gt
= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s
= 50.01 m/s - 110.64 m/s
= -60.63 m/s
Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.
The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)
= √((-60.63 m/s)² + (72.77 m/s)²)
= √((3676 m²/s² + 5295.48 m²/s²)
= √(8971.48 m²/s²
= 94.72 m/s
The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°
So the impact velocity is 94.72 m/s at -39.8°
c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m